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BIBLIOGRAPHIC RECORD TARGET
Graduate Library
University of Michigan
Preservation Office
Storage Number:
ACV0278
UL FMT B RT a BL m T/C DT 07/19/88 R/DT 07/19/88 CC STATmmE/Ll
035/1: : | a (RLIN)MIUG86-B35983
035/2: : j a (CaOTULAS)l 60645772
040; : | a MiU | c MiU
100:1 : I a Durell, Fletcher, | d 1859-
245:00: | a Plane and solid geometry, | c by Fletcher Durell.
260: : | a New York, | b Charles E. Merrill co. (c 1911 [cl904]
300/1: : |avi, 7-552p. | b front, (group of ports.) diagrs. |cl9cm.
490/1:0 : | a Durell's Madiematical series
650/1:0: | a Geometry
998: : |cWFA |s9124
Scanned by Imagenes Digitales
Nogales, AZ
On behalf of
Preservation Division
The University of Michigan Libraries
Date work Began: _
Camera Operator: _
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PLANE AND SOLI!
GEOMETRY
BY
FLETCHER DUEEI.L, Pii.D.
NEW YORK
CHARLES E. MERRILL CO.
44-60 East Twentv-thtrd Street
1911
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DuroiPs r^Iathf
Plane Geometry
341 pages, 12mo, hiiii Icatlu-r
Solid Geometry
213 pages, 12nio, lialf leatliev
l^lane and Solid Geometry
614 pages, 12ino, half leather
I'lane Trigonometry
184 pages, Svo, oloth . . .
Plane Trigonometry and Tallies
298 pages, Svo, uloth . . .
P|ane Tiigonometry, with Surveying
Tables
In preparation
Plane and Spherical Trigonometvy,
Tables
351 pages, Svo, cloth ....
Plane and Spherical Trigonometry,
Surveying and Tahles
In preparation
Logarithmic and Trigonometric Table;
114 pages, Svo, cloth ....
Copyright, 101)4, by Cliarles E. Merrill Co.
[5]
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PREFACE
One of tlie main pui-poses in wi'itiug this book has
been to try to present the subject of Geometry so tliat llie
pupil shall Tiuderatand it not merely as a series of eorrei^t
deductions, but shall realize the value and meaning of its
principles as well, This lapect of tlie subject has been
directly presented in some places, and it is hoped that it per-
vades and shapes the presentation in all places.
Again, teachers of Geometry generally agree that the
most difficult part of their work lies in developing in
pupils the power to work original eseixsises. The second
main purpose of the book is to aid in the solution of this
difiieulty by arranging original exercises iu groups, each
of the earlier groups to be worked by a distinct method.
The pupil is to be kept woi-king at each of these groups
till he masters the method involved in it. Later, groups
of mixed exercises to be worked by various methods are
given.
In tlie current exercises at the bottom of the page,
only such exercises are used as can readily be solved in
connection with the daily work. All difflcalt originals are
included in the groups of exercises as indicated above.
yimilarly, iu the writer's opinion, many of the numeri-
m
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cal applications of geometry call for special metliods of
solution, and the thorough treatment of such exercises
should be taken up separately and systematic ally, [See
pp. 304-318, etc.] In the daily extempore work only such
numerical problems are included as are needed to make
clear and definite the meaning and value of the geometric
principles considered.
Every attempt has been made to create and cultivate
the heiiristic attitiiile on the part of the pupil. This has
been done by the method of initiating the pupil into
original work described above, by qiieries in the course
of proofs, and also at the bottom of different pages, and
also by occasional queries in the course of the text where
definitions and discussions are presented. In the writer's
opinion, the time has not yet come for the purely heuristio
study of Geometry in most schools, but it is all-important to
use every means to arouse in the pnpil the attitude and
energy of original investigation in the study of the siihject.
In other respects, the aim has been to depart as little
as possible from the methods most generally used at
present in teaching Keometi-y.
The Practical Applications (Groups 88-91) have been
drawn from many sources, but the autliur wishes to ox-
prcKs his especial indebtedness to the Committee which has
collected the Real Applied Problems published from time
to time in Sfhool Science nnd Miitheinatii.-x^ and of which
Professor J. F. ilillis of the Francis "W. Parker School
of Chicago is the chairman. Page 360 is dne almost en-
tirely to Professor AViiHam Eetz of the East High School
of Koehester, N. Y.
FLETCHER DURELL.
LiwaBNCEyiLi.E, N. J., Sept. 1, 1904,
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TO THE TBACHEK
1. In working original exercises, one of the chief dif-
ficulties of pupils lies ill their inability to construct the
figure required and to make the particular enunciation
from it. Many pupils, who are quite unable to do this
preliminary work, after it is done can readils' discover a
proof or a solution. In many csereisea in this book the
figure is drawn and the particular enunciation made. It
is left to the discretion of the teacher to determine for
what other exercises it is best to do this for pupils.
2. It is frequently important to give partial aid to the
pupil by eliciting the outline of a proof by questions such
as the following: "On this figure (or, iu these two tri-
angles) what angles are equal, and why?" "What lines
are equal, and why?" etc.
3. In many cases it is also helpful to mark iu colored
crayon paii's of equal lines, or oE equal angles. Thus, in
the figure on p. 37 lines AB and DE may be drawn with
red crayon, AO and DF with blue, and the angles A and J>
marked by small arcs drawn with green crayon. IE
colored craj"ons are not at hand, the homologous equal
parts may be denoted by like synibois placed on them,
tllUS : F
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^ TO THE TE.U'HER
(either given or to be proved) by lifi-
ures denoting the onk^r in which 11.,/
lines are to be taken. Thus, if 0.1:
0C= 01) OB, the relation iimy !)e
indicated as ou the figure.
4. It is sometimes helpful to vaiy
the symbolism of the book. Thus, in dealing
equalities a convenient symbol for "angle" is 2)^,
aa ^ A > 2^1 B.
5. Bach pupil need be required to work only so many
originals iu each group as will give him a mastery of the
particular method involved. A large number of exercises
is given in order that the teacher may have mnuy to select
from and may vary the work with successive classes.
6. It is important to insist that the solutions of exer-
cises for the first few weeks be carefully written out; later,
for many pupils, oral demonstration will be suiilcieiit.and
ground can be covered more rapidly "^y its use.
7. In leading pupils to appreciate the meaning of
theorems, it is helpful at times to point out that not every
theorem has for its oljject the demonstration of a new and
unexpected truth (i, e., not all are "synthetic"), but that
some theoreips are analytic, it being their purpose to re-
duce an obvious truth to the certain few principles with
which we start in Geometry. Their function is, therefore,
to simplify and clarify the subject rather tbau to extend
its content.
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TABLE OF CONTENTS
Definitions akd Fiiwr Piunciplks . .
Book I. Rectilinear Fh;uhes ....
Boos II. The Cihclb
E(.aK III. PKOPORTiaS. SIJKI.AU PoLVt
Book IV. Aki:a i-i' PoLvi;(.Na , . , .
Book V. Eeri-l.^r Polyiions. Mkahurf
KuiiDiutjAL Applications or I'lanl V.\:f
Book VI. Likes, Planes and Anolks
BofJK Vil. l'OLVHKI)liO\8
Booii VIH, CVE,INDK1:S A.SD C'o.NLiSi .
Book IX. The Si'iiiiitE
MojiEincAi. Esn;RC(SEs in Somd Geomi
MoDEiis (JEOJfETiiic Concepts ....
IIlSTOHV OF UHOMl-.THV
Review E.'iEiicisi'is in Pi.ani; tlr.oMTyn
Review E,\-ehcisi;s in Solid (iEoMEii:
I'R.vrncAL Api'i,icatio\s of Ceomf.ti
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SYMBOLS AND ABBREVIATIONS
+ plus, or increased hg.
Adj. . .
adjareiit.
— minus, or dimimshed hj.
Alt. .
alternate.
X multiplied hj.
Art. .
article.
-i- divided hy.
Ax,. .
ax>«m.
~ equals; is (or arc) equal lo.
Coiistr.
conslrucliox..
A approaches {as a liiiiit] .
Cor, .
corollary.
K= is (or are) equivalent to.
Def. .
dcfimtion.
> is (or ace) greater than.
Ex.. .
exercise.
< is {or are} less than.
Est. .
exterior.
.: there/ere.
Fig. .
fignrc.
X perpejidiCKlar , pcrpeiKUciihir t
, Hyp. .
hypothesis.
or, is perpendicular Id.
Ident.
identity.
Ii.t. .
interior.
it parallel, or, is parallel lo.
I'oBt. .
Ila parallels.
Prop. .
. propositioH.
I , i angle, angles
m. . .
right.
A, A triangle, triangles.
Sns. .
. suggestion.
CJ , E£! parallelogran}, paralklogram.i
Sup. .
supplemciila
O, ® circle, circles.
St. . .
. sfmigbt.
q. E. D, quod erat demonsfrandum
that is, w
hich Wiis 10
proved.
Q. E. F. quod erat faeiendum; that i
, which was
to be ma^ie.
A few other abbreriatione and symbols will be introduced i
their meaning iodicaied latei: on.
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DEFINITIONS AND FIRST PRINCIPLES
IKTRODOCTORY ILLUSTRATIONS. DEFIHITIOWS
1, Computation of an area, Pr;ieticiil experience lias
taught men that cei'fcaiii ways of dealing with objects in
the world about us are more advuiifageons tlmn others;
thus, if it be desired to find the num-
ber of square yards in the area of a
floor, we do not mark off the floor
iuto actual square yards and count
the number of square yards thus
made, but pursue the much easier
course of measuring two lines, the
length and breadth of the floor, obtaiuicg 7 yards, say, as
the length, and 5 yards as the width, and multiplying
the length by the breadth. The area is thus found to be
35 squai'e yards.
Let the pupil determine the area o£ soioe couveuient floor iu oauli
of these two waj-s, aud compare the inbof of the two processes.
2. Computation of a volume. Similarly, for example,
in order to determine the n amber of cubic feet which a
box contains, instead of filling the box with block.s of
■Wood, each of the size of a eubie foot, and
counting the number of blocks, we pur-
sue the much easier course of measuring
the number of linear feet in each inside
edge of the box and multiplying together
the three dimensions obtained; thus, if
19)
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]0 PLANE GROHrETE"/
tlie inside dimonMioii,-: an: », ■! anil 3 fci-'t, llif volumu is
5 X 4 X 3, or CO cubic tuat.
Similarly, the direet methoii of mensuriiif! the nuniTjer ii£ bushels
oE wheat in a bin is to fill a bushel meaaure with wheat from thu
bin, time after time, tiH the biu is exhausted, and count the number
of times the bushel measure is used. But a much leas laborious
method is to measure the three dimensions of the bin in inches
and divide their pvoduet by the number of eubic inches in a buabel.
Let tlie pupil in like manner compare the lal)or of finding the
number of feet of lumber in a given bloek of wood by actually saw-
ing the Mock np into lumber feet, with the labor of measni'lng the
dimensions of the block a7id computing the number of lumber feet
by taking the product of the linear dimensions obtained.
3. Unknown line determined from known lines. Tlie
fituiloLit is also probably familiar with the faut tluit, ii.v
comjiutiitionB based on the relations of cer-
tain lines whose lengths are known, the
lengths of other unknown lines may be
determined without the labor of measuring
these unknown lines.
Thus, for instance, if a ladder 5 yards long lean
agninst a wall and have ita foot 3 yards from the
wall, the hoight of the top of the ladder from the
ground may be determined thus:
(Height)^ = 5'^ _ 3' = 23 — 9 = IG.
4. Economies in representing surfaces, lines, etc. Other
principles of advantage of an even more general eharaeter
occur in dealing with geometric objects. Thus, since only
one straight line can be passed through two given points,
tlie two points may be taken as a highly economized
symbol or representative of the line, by the use of which
much labor is saved in dealing with lines, and new results
are made attainable.
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riT.nMV.TIilf JtAnNITUDES 11
Similarly, siiieo only one fliit, or plane surf:iee enn be
passed through three given points (not iu the same straight
line), these three points may be taken as a symbol or
representative of the tlat sni-face. This gives the advan-
tage not only of reducing an nnlimited surface to three
points, but also of giving for the plane a symbol made up
of three parts. By varying one of these parts and not the
others, tlie plane iriiiy be varied in ouo respeet and not in
others; also planes having certain properties in <;onimon
may he grouped togetlier, and dealt with iu the groups
formed.
5. Geometry as a science. Definition. The above illus-
trations serve to Amw limv advantageous it often is to deal
with geometric magnitudes by certain methods rather than
others.
In the study of Geometry as a science we proceed to
niake a systematic examination of these methods.
Geometry is the science which treats of the properties of
continuous magnitudes and of space.
GEOMETRIC MAGNITUDES
6. Solids. A physical solid is a portion of matter, as a
bioek of wood, an iron weight, or a piece of marble.
The portion of space oceiipied by a physieal solid may
be considered apart from the physieal solid itself, hence
A geometric solid is the portion of space occupied by a
physieal solid, or definitely determined in any way. Hence,
also, a geometric solid im a limited y)ortiou of space.
One advantage id UKing sEOmetrie solids liua in tlda. If we deidt
with phyaioal Bolida oiUy, aa blocks oE wood, uuurMu, iron, etc., we
should need to determine tlie jiroperties oJ each kind o£ physical aoiid,
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]2 riiANE r.KOMF.TKV
separately; but, by deterrainlug tiio jiropertits of a geomotric. nolitl,
iwe deteimine onee for all the pvoperties of evoiy physical solid, tio
matter what its material, that will exactly fill the space occupied by
the giveu geometric solid.
Hereafter in this book the term "solid" is understood to
mean geometric solid, naless it be otherwise specified.
7. Other geometric magnitudes defined as boundaries,
A surface is the bonudiiry of a «oiid.
A line is the boundary of a surfaiie.
A point is the boundary of a line.
The solid, surface, line, and point are the fiiijdr.mcntal
geometric magnitudes,
8. Geometric magnitudes defined by their dimensions.
A solid has three dimensions; viz., length, breadth and
A surface has two dimensions, length and breadth.
A line has one dimension, length,
A point has no dimension. Hence a point is that which
has position, but no magnitude.
9. Geometric magnitudes defined as generated by motion.
Aline is that which is generated by the motion of a point .
A surface is that which is generated by the motion of a
line (not moving along itself).
A solid is that which is generated by the motion at a
surface (not moving along itself).
The three independent motions by which a solid is generated ilhia-
10. Geometric magnitudes as intersections. The inter-
section of two surfaces is a line; of two lines is a paini.
It is sometimes more advantageous to regard geometric
magnitudes from one of the above points of view, some-
times from another.
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LINES 13
11. A geometric figure is any combination of poiots,
lines, surfaces, or solids.
12. The form or shape o£ .i figure is determined by the
relative position of its parts.
13. Similar geometric figures are those whieh have the
same shape.
Equivalent figures are tliose which have the same size.
Equal or congruent figures are those which have the
same >jli<tpr and .■<izt\ iind can, therefore, be made to
coincide.
14. A point iri represented to tin- eye by a dot nnd ie;
niimed by ll lett.et- Liftixed to the dot, as the point A, -A.
15. A straight line is a line such that, if any two points
in it be fixed and the line rotated, every point in the line
will retain its original position.
A straight line is also sometimes described as a line
which has the same direction throughout its whole extent;
or, as the shortest line connecting two points.
The word "line" may be used for — - — ■-
"straight line," if no ambiguity results.
16. A curved line is a line no portion
of which is straight. The word "curve "
is often used for "curved line."
17. A broken line is a line made up of different straight
18. A rectilinear figure is'a figure composed only of
straight lines; a curvilinear figure is a figure composed
oTily of cnrved lines; a mixtilinear figure is a figure
containing both straight and curved lines.
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.14
PLANE GI,OMKTr
19. Kinds of straight lir.e. A .siraiKlit lin.-
dcliTiite Of iudolinitc in Iciij- h.
Tho line of (Jefinili^ k'li ;th is Hoiurtiincs b
sef/inmi ot- m-cl.
Otb^r kiiuta .>f RlraiKht line are .li.fiiicd in Arts. 2C an.
20. Naming a straight line. A straight line i
by naming two of its points, as the
line AB (a sect) ; or the line CD (in-
definite in length). A segment or sect
may also be denoted by a single
{small) letter, us the Hne a. J" ~
21. The circumference of a e
of whieh id equally dirilaiit fro
called the center.
-de
line every
i point w
ANGLES
22. An angle i» the amount of opening between two
sti'aight linos which meet at a point.
The sides of an angle are the lines whose intersection
forms tlio iin^jli'; the vertex of an angle is the point in
which the sides intersect.
23. Naming angles. (1) The most pr
naming an angle is to use three let-
ters, one for a point on each side of
the angle with the letter at the vertex
between these two, as the angle ABC.
SiDce the aiae of an angle is indepen-
dent of the length of its aides, the points
nnmed on its aides may be taken at any
plauB on its siilas. Thus, the angles AOl),
BOD, iSO£,dOFa.ve all the same auglo.
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ANGLKfi
(2) In fiasc l.liei-c is Init onr iiiigk" at a eivoJi ven
the lettor iit the vm-tox alone is sufri«ient to denote
angle, as the angie in the last figure.
(3) Sometimes a letter or figure phiced inside ,
the angle and near the vertex is a convenient ziL
symbol, as the angle «.
24. A straight angle is an !ii\<;li
same straight Hue, hut which ex-
tend in opposite direetions fi-oni
the vertex, as the ang'le BOA.
25. A right angle is one of two eniiiil
one Ktr;tight liiiu hKiuliiig' ;uioll
line I'Q meetw lino AB so a.^ to uial'ie
angle PQA equal to angle I'QIi, iVM-h of
these angles is a right angle. A riy^lit
angle is also liiilf of a straight angle.
whose
sides li.' in the
^
"
ef|niil .1
-ill lint
nffles made hy
. Thus, if the
26. A perpendicular is a line th
makes a right angle with a given line;
thus PQ in the last figure is perpendicular to liA. The
foot of a perpendicular is the point in which the perpen-
dicular meets the line to which it is drawn, thus Q is the
foot of the perpendicular PQ.
27. An acute angle is an angle
less than a right aiiijle, as the
angle ^icr.
28. An obtuse angle is a
angle greater than a right angl
hut less than a straight angle, l
angle A OH.
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IG
I'LANE GEOMETP.V
29. A reflex angle is an angle
greater than a straight angle, but. less
tlian two straigiit angles, as angle ^ Oi^.
Ill tliU book, angles larger tliiin a atraigUt
angle are not considered unless special mention ia ni;t.i
30. An oblique angle is an angle tliiit is
nor a straight iiiigle. Hence, oblique angle is
for acute, obtuse, and reflex angles.
31. Adjacent angles are angles
whii^li have a common veiiex and a
common side between them, an anglt's
AOBan<iBOa.
32. Vertical angles are angles
which have a common vertex and
the sides of one angle the prolonga-
tions of the sides of the other angle,
as the angles AOG and BOD.
53. Complementary angles are two
gotlicr equal a right angle, an the angles
AOP a\>d rOQ.
Hence, the complement of an angle
is the difference between that angle
and one right angle.
u~
54. Supplementary angles urn two
angles whicii together equal two I'ight /P
angles (or a straight angle), as the an- y'
gles A OP and POS. / _^
Hence, the supplement of an angle '^
is the difference between that angle and two right
angles.
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ANai.ns 17
85. Angles as formed by a rotating straight line. If
the line OB start in the position OA and rotate to the
position OjB, it is said to generate tlie
LAOB. ?
The size of an angle may, therefore, \ I _,-'
be considered as the amoniit of rotation ^ _ '^.j, .-'^ ^
of a line about a point, from the origi- /^
nal position of the hue. p/
If the rotation ia eon tinned far
enough, a right angle {LAQG) is formed; afterward an
obtuse angle {lAOD); then a straight angle {LAOli),
and a reflex angle ( lAOV), etc.
An advantage of this method of forming or coneeiving
angles ia that by continning the rotation of the moving
line an angle of indefinite size may be formed.
36. Units of angle. A right angle is a unit of angle
useful for many purposes. Sometimes a smaller unit of
angle is needed.
A degree is one-ninetieth of a right angle; a minute is
one-sixtieth of a degree; a second is one-sixtieth of a
niinnte.
BesidoB tbuse, oDjuv units of angle are used for ceitain puipoaes.
SURFACES. DIVISIONS OF GEOMEXEY.
PARALLEL IINES
37. A plane is a surface such that, if any two points in
the surface be joined by a straight line, the straight line
lies wholly in the surface.
Hence, it plane figure is a figure sueh that all its points
lie in the same j)laiif.
38. A curved surface is a surface no part of which is
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18 I'L.VXK f3E0M]':'Jliy
39. Plane Geometry is thnt. bram^li of Geometry wljieli
treats of plane figuves.
40. Solid Geometry is that branoli of Geometry wLiiih
treats of figures all points of whieh are not in tiie samei
plane.
41. Parallel lines are straight lines in tde s;ui;o piitno
Tvliicli do not ]neot, liowever fur they be protUiood.
Ex. 2 Which of the capital letters o£ the alphabet are str.iight,
which curved, which broken, which curved aud straight linta
coBihiiied t
Kl. 3. Draw an acute angle. An obtuse angle. A refliix angle.
Ex. 4. Draw two adjacent angles. ' Two vertical angles.
Ex.6. What is the complement of an angle of 43"! What is its
Bupplemtnt f
Ex. 6, What is the complement of 57° 19' 1 of 62° 23' 43" 1 What
is the supplement of each of these !
Ex. 7. At two o'eiocli, what k the angle made by the hour an.l
minute hands of a clock 1 at three o'clonk f at five o'.'lo^-k ?
Ex. 8. At 1 ;30 o'clock, what anf;lo do the hands of a clock raak,' ?
at L>:I5t at 8;-lj?
Ex. 9, What kind of an angle is the supplement of . an obtuse
angle; of an acute angle f of aright angle t
Ex. 10. What kind of a surface is the floor of a roomT the surface
of a baseball? the surface of an eggt tlie surface of a hemisphere!
Ex.11. If the surface JBCD n
the right, what solid is generated h
itt What surfaces are generated
bounding lines T What
tlcos of its angles t
;nerated by ^~~S\ '
ated by its 1
by the vet- Jr=^-:."v.:::
■II
---=.>a
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GKOMETKIC MAGNITUDES ISI
Ex. 12. Draw two supple me ntavy adjacent iicgles. Also two sup-
plementary angles that are not adjaeent. Alao two adjacent angles
that are not supplementary.
Ex. 13. The sum of a right angle and an acute angle iswliat kind
of an angle? Their difference is what kind of an angle!
Ex. 14. The sum of an obtuse angle and a right angle is what
kind of an anglel Their difference is what kind?
Ex. 15. If three straight lines meet (but do not intersect) at a
point in a plane, how many angles have this point as their common
vertes? Draw a figure illustrating this and name the angles,
Ex. 16. How many angles are formed if four lines meet ibut do
not intersect) at a point in a plane!
Es. 17. How many degree
■s art' ii
lan auglo whiph equals tw
oomj.lement!
Ex. 18. How many degret
^s in ai
a angle which equals ouf
its supplement!
Ex. 19. What kind of an
angle i
s greater than its suppk
PRIMARY RELATIONS OF GEOMETRIC MAGNITUDES
42. C<!vtjuii primary relations of geometric objects have
already been giveu in the defiiiitioiis uaod for geometric;
objecte. We now proceed to investigate the relations of
geometrie maenifcudes more generally and systematically.
43. An axiom is a troth accepted as requiring no
demonatratioti.
44. Two kinds of axioms are used in geometry:
1. General axioms, or axioms which apply to o.'-.her
Ivinds of rinaiitity as wi>ll as to geometric magiiitudesj for
instance, to numbers, forcca, masses, etc.
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liU PLANK GEOIIETHY
2, Geometric axioms, or iisioms \ylii(!h apply to gee-
metrii; tiiaguitiidtis alone.
45. The geEcral axioms may be stut.t'd as follows;
GENERAL AXIOMS
1. Things H-hieh arc rqiud to fJir siime thintf, or to rqual
ihingf, en' rqiiuJ fo Hwli nthrf.
2. Jf rqnah be ciddfid to pcinalu, ihe niim^ are equal.
.■J. // equals he subtracted from equals, the remainders
are equal.
4. Doubles of equals are equal; or. in general, if equals
he multiplied hy equals the pndue/s tire eiii'ul.
r>. Halves of equals are equal; or, in general, if eqnah
he divided by equals the quotients are equal.
6. The whole is equal to the sum of its parts.
7. The whole is greater than any of its parts.
8. A quaiility may he substituted for its equal in any
proeess.
9. Jfeqiials be added to, or suhtr acted from, miequals, the
results are unequal in the same order; if uiiequals be added
to Uiiequals in the same order, the results are unequal in
that order.
10. Doubles, or halves, of uuequals are unequal in the
same order,
H. If miequals he subtracted from equals, the remainders
are unequal in the reverse order.
12. If, of three quantities, the first is greater than the
second, and the second is greater tJtan the third, then the
jirsi is greater than the third.
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GEOMETKIO AXIOHR 21
46. The value of the general axioms. Tlie axioms given
above seem so obvious that the student at first is not likely
to realize their value. This value may be illustrated aa
follows:
II the distnnee fvom Washington to PhiladelphiB be known, and
also the distance fvom Philadelphia to New York, the distance from
Washington to New York may be obtained by adding together the two
distaneea named; for, by axiom 6, the whole is equal to the sum of
its parts. Thus the labor of actually meusuriiig the distauee from
■Washington to New York is savod.
Agaiu, if the lu'ifrht ot a suhoolboy o£ a ^'ivon age in Paris be
measured, and the height of a like schoolboy in New York be meas-
ured, and the result o£ the measnrempiit in each case is the same, we
know that tho boys are o£ the same height, without the labor and cost
of briugiugtlieboystogether and comparing their heights directly ; for,
by axiom 1, things which are equal to the same thing are equal to
each other.
Thus the general axioms are to be looked at not merely
as fundamental equivalences, bitt also as fundamental
economies. For many purposes the latter point of view is
more important than the former.
47. Tlie geometric axioms may be stated as follows:
GEOMETRIC AXIOMS
1. Through Uvo given poinh onhj one slraighf line can
be passed.
2. A geometric figure may be freclij moved in space with-
out any change inform or she.
This aiiom is equivalent to regarding space as mu/orm, or homn-
loidal; that is, as having the same properties in all its parts. It has
already been assumed in some of the deSnitions given. See Arts. 15
and 3u.
3. Through a given point one straight line a'nd only one
can be drawn parallel to another given straight line.
By Art. IS, geometric figures which coincide are equal.
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22 PLANE GEOMETCr
48. Utility or uses oi the geometric axioms. By means
of the firet geonintric axiom we urc able to ishrink or con-
dense any striiight line into two points. Later tiiis atlvan-
tage gives rise Lo many other advantages. By tlie second
geometrie axiom, tbe knowk'djfe whicli ive liaye of ono geo-
metric object may be transferred to anotlier like object, how-
ever widely separated in space. The utility of the third
geometric axiom can be miwle more evident wlieu we come to
use it in proving ]iew geometriu truths.
49. A postulate in geometry is a oonstruction of a geo-
metric figure admitted as possible.
50. The postulates of geometry may be .-stated as follows :
1. Through any two points a straight line way he dt-airn.
2. A straight Unetnay ie extended indejiHitelij , or it maij
be limited at any point.
3. A circumference may be described abont any given
point as eenier, and with any given rifdiii.'^.
These postulates limit the pupil to tbe hsb of the slrnight-eciKed
ruler and tlie compasses in eonatmoting Gbuibb in geometry. One of
the objects of the study ot geometry is to disi'over what geometrin
figures can be coustructed by a combination of tbe eleiiienfnry eon-
atraetions allowed in the postulates ; that is, by the use of tho two
simplest drawing instraments.
51. Logical postulates. Besides the postulates which are
used in the actual construction of figures, there are certain
other postulates which are used only in the processes of
reasoning. Thus, for purposes of reasoning, a given
angle may be regarded as divided into any convenient num-
ber of equal parts. Whether it is possible actually thus to
divide this angle on paper by use of the ruler and com-
passes, is another question.
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DEMONSTRATION OF PROPEItTlER '26
DEMOKSTKATION OF GEOMETRIC RELATIONS
52. A geometric proof, or demonstration, is a course of
reasoning by whidi a relation lietween geometric objects ia
established.
53. A geometric theorem is a statement of a truth con-
cerning geometric objects which requires demonstration.
Es. The sum of the angles of a triangle equals two ri);'it angles.
54. A geometric problem is a statement of the construc-
tion ot a geometric iigiire which is required to be made.
Es, On a given line to eonstruet a triangle toutalning three equal
65. A proposition is a general term for either a theorem
or a problem.
Thus, propositions are sulidiviileil into two classes:
1, Theorenas; 2, Problems.
56. Immediate inference is of two kinds:
1. Changmg the point of view in a given statement.
Thus the statement, "two straight lines drawn through
two given points must coincide," may be changed to "two
straight lines cannot inclose a space."
2. Reasoning which involves but a single step.
Ex. "All straight angles are equal;"
.'. "All right angles are equal." (Ax. 5.)
57. A corollary is a truth obtained by immediate infer-
ence from another truth just stated or proved.
58. A scholium is a remark made upon some particular
feature of a proposition, or upon two or more propositiona
which are compared.
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2-i PLANE GEOMETRY
59. Hypothesis and conclusion. A |iroiiosition consists
of two parts :
1. The hypothesis, oi- Unit wlikh is known or granted.
2. The conclusion, or that which is to he ]ii'ovi'd oi- coii-
stvueted.
Thus, in the pfoposition, "if two straight liiicH are per-
pendienlar to the same line, they are parallel," the
hypothesis is, that two given lines are perpendienlai" to
jinother given line; the eoncUision is, that the two given
I'mes are parallel.
60. The converse of a pvoposition is another proposi-
tion formed by interchanging the hypothesis and the
conclusion of the original proposition.
Thus, theorem, "every point in the perpendicular bisector of ii. ]itie
18 equidistant from the estremitiea of the line;"
Cotirefse, "every point equidiatant from the extremities of aline
lies in the perpendicular biaeetor of the Hue."
Or, in geneml, theorem, "if A is B, then A' is F."
en»Terse, " if X is F, then A is nr'
Frequently a converse is formed by interthanfiing p^i-t only of an
hypothesis with part or all of the eoneluuion, or vice vi^rsii.
Thus, theorem, "if ^ is B and C ia D, then J is I'r'
com-erse, "" if J is B and X is Y, then C ia Jh"
The converse of a theorem is not necessarily trne. Tims,
it is true that all right angles are equal, but it is not true
that all equal angles are right angles.
61. The opposite of a theorem is a theorem formed by
making both the hypothesis and the conclusion of the
original theorem negative,
Ex. theorem, "if A is B, then X is Y;"
opposiff, "if A is not B, then X is not 1'." '
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DEMONSTRATION OF PROPEETIES 25
If a direct theorem and its opposite ace both true, the uonverge is
known to bo true without proof.
Also, if a theorem and its converse are both true, the opposite ia
known to be true withont proof.
Thus certain economlea arise in the demonstration of theorems.
62. Methods of geometric proof. Sevuciil principal
methods of proving theorems are used in geometry.
1. Direct demonstration.
2. Proof by superposition, in which two iijjureiS are proved
equal by makiiii^ one of Ihein eoiiieide wifh the other.
3. Indirect demonstration, wliieh eonsists essfniially in
showing that a given statement is true by i^howiiig that its
negative cannot be true.
Other special methods of proof will be pointed ont as
they occur in the coarse of the work.
63. FornI of a proof. The statement of a theorem and
its proof eonsist of certain distinct parts which it is impor-
tant to keep clearly in mind. These parts are;
1. The general enunciation, which is the statement of
the theorem in general terms.
2. The particular enunciation, or statement of the
theorem as applied to a particular figure used to aid the
mind in carrying forward the proof.
3. The construction of supplementary parts of the figure
(not necessary in all proofs).
4. The proof. This must include a reason for every
statement.
5. The conclusion.
The letters Q. E. D., standing for "quod erat demon-
strandum," and meaning "which was to be proved," are
usually annexed at the end of a completed demonstration.
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PLANE GEOMETi;V
EXERCISES. CtiOi
Es. 1, InthofiKurempas-
ure AB: then meivsure BC. A — _— -a
Now find AC without meas-
uring it. Wliat axiom hiive you used 1
Ex. 2. If /L ^0^ = 00°, I BOC=m'' and ■'; /'''
ZCOi) = 130°; find without measui-ing them a ~X /
AOC and BOD (reHe.";). What axiom have you N, /
Ex. 3. Prove by repeated use of the first
part of Asiom 1 that magnitudes equal to equal [
magnitudes are equal to each other; (thus, I>
given J=!.r, li=y, and x~'j. Prove A^B).
Ex. 4. Give a namevipal iliuKtration of Axinm 11,
Ex. 5. Show liy ttia axioms that a part is equal to a whnlo di-
minished by the remaining part,
Ex. 6. Show that Axiom 1 is a special chsb of Axiom K,
Ex. 7. It a = x + }j and ;c = y, show by use of the axioms that
Ex. 8. Di'aw a line and produce it so that the produced part shall
equal another given line.
Ex. 9. By use of the eompasaes, mark off on a given line a part
twieo as long as another given line.
Ex. 10. On a givenlinemarkoff.byfoive^t uses of theoompHSses,
a part four times as long as another given liue.
Ex. 11. Di^aw three straight lines and denote them hy I, ra,
andii. Then draw a line l + ni-~n, and also a line I — 2'i(+3ii.
PROPERTIES OF LINES INFERRED IMMEDIATELY
64. If two straight lines have two points
the lines coitidde throughout their whole extent (Art, 47,
Geom, As. 1).
Hence, two straight tines can intersect in hut one point.
65. If two straight lines coincide in part, they coincide
throughout,
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PEOPERTIES OP ANGLES 27
66. Only one straight line can be drawn eonnecHng two
given points.
67. Two straight lines cannot enclose a surface.
68. A given straight line {sect) can he divided into two
equal parts at but one point.
For (by Ax. 5) halves of equals (or of the same thing)
are equiil.
PROPERTIES OF ANGLES INFEEEED IMMEDIATELY
69. All glrnight angles arr equal.
70. A straight angle can he dividfd into lieo equal
angles by but one line at a given point .
in the given straight Zme.
For (Ax. 5) halves of the same mag-
nitude are equal.
71. Hence, at a given point in a straight line hut one
perpendicular can he erected to the line.
72. All right angles are equal.
For all straight angles are equal (Art. 69) and halves
of equals are equal (Ax. 5).
73. The sum of the two adjacent angles /
formed by one straight line meeting an- F^i
other straight line equals ttvo right angles.
For the angles formed are supplementary adjacent
angles (Arts. 31, 34).
74. If two adjacent angles are together equal to a
straight angle {or two right angles), their exterior .Hid«'ts
form one and the same straight line.
For their esterior sides form a straight angle, and
hem;e must lie in a straight line (Art. 24).
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28
}'T.ANE 03?,OJIETRY
75. The cowplemeiiix of Iwo equal nn^jlr-: /nr equal
(Art, 72 and Ax. 3); the aiipplemetits
of two equal angles are equal (Art. Gi),
Ax. 3).
76. The sum of all Ike. oiujles about a
point equals four rhjhl amjleti.
Thus, la + lh + lc + ld^-le--^
4 rt. A..
■77. The SUM of all the angles about
a point on the same side of a straight
line passing through the point equals
two right angles.
Thus, Zj3 + Z5 + Zr = 2 i-t. i.
EXERCISES. CROUP 3
Ei. 1. How many different straight linps are determineii liy tliree
points not in the Same straight line?
Ex. 2, How many strniifht lines are determined by fouc points in
a plane, no three of them being in the same stra:(;(it line ?
Ex. 3. If, iii Fig. 2 above, Aa,b,c. cl^ 40°, 50°, 60°, 70° i-esppp-
tively, find Z e.
Ex. 4. IE,- in Fig. 3 above, the lines forming the an^ld q are per-
pendieulat to each other and I p = 47°, find the other angles of Iho
fignre.
Ex. 5. Measure Zrt o£ Fig. I on preceding page. Find Zb with-
out measuring it. Now measure Ih and
compare the two results, , A
Ex. 6. Given QB1AH, PB 1 BC, and \ ,q
IABC = 1W; find tha otliot angles of the \ ^^
figure. \l rrr\'
B a
Ex. 7. Arrinpe file points in ^ pKne
so that the few ! st n imber of st light line'! i av pass through them,
no line to pass through more than IhrfiL [ lul?
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PLANE GEOMETRY
Book I
RECTILINEAR FIGURES
Proposition I. Theoreji
78. If one simight line inUrsecfa anotlipr siraight line,
the opposite or vertical angles are equal.
GiveE the straight linos AB and CD iutersoetiiig at
the point 0.
To prove ^AOC= IDOB and IA0T)= ICOli.
Proof. /. AOG+ ZA0D=2 rt. A, Art. 73.
[tkv Kiim (jf (ICO ndjaecut angles formed hij one striikjht hue meclimj
aiiullicr straight line equals tu-o riylit angles).
Also ZfiOD+ZJ.OD==2rt. A,
isame reason).
.-. ZAOC+^AOD=^lBOI>+ZAOD, A^i. i.
{things equal lo the same ilibuj are eqiiul to t'udt other).
Subtracting ZAOD from the two equals,
/.AOG = lBOD, Ax. 3.
[ifeijiials he siiblraotcd from equals, the remainders are equal).
Id liliu iiiamu;r it may be proved that
IA0J) = IC0B.
O.B. B.
Ex. If i
without uihu
n the above 6gure IDOB^IQ".
isuriug them.
Cud the other angles
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30
BOOK I. PLAXE GEOMETltV
Proposition; II. Tiieoeem
79. If, from a point in a perpendicular' to a given line,
two oblique lines be drawn cutting off on the giceu line equ<tl
sffimenia from the foot of the perpendicular, the oblique lines
are equal and make equal aiujle.s with the perpendicular.
Given a line AB witli CI) ± to it at the point C. and
FR and FQ drawn from :iny point as P in CD, cutting off
CR = QCon AB.
To prove FR ~ FQ and l CFR = l CFQ.
Proof. Fold over the iigiive DCB abont J}C as an axis
till it i;omes in the plane DOA. Geom. Ax. 2.
Then ^DCB = /. DCA {allrighl A are = ). Art. T2.
.■. line CB will take the dii-eetion of CA.
But
Hencf
And
CJi= CQ.
.: point R will fall on point Q.
ine PR will coincide with line FQ.
Hvp.
■u-aiyia
ectiu'j U
I CFR will coincide with / CFQ.
.: FR = FQ, and Z CPE = I CFQ,
{geometric figures which coincide are equal).
Ex. 1. Point out the hypothesis and the oonelusioii ii
eounciatton of Prop, I. Also point thorn out in ti
enunciation. Do tho same tor Prop, II.
Ex. 2. If three straight lines intersect at a point, 1
the angles formed is it necessary ti
all the augleat
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LINES AND ANGLES 31
pROPOsiTioK III. Theorem
80. From, a given point wUhftut a straight line but one
perpendicular can be drawn to the line.
Given the str,iif;iit lino AB, P any point witlioiit: AP,,
PQ ± .1 /i, iiud PR any othet- line dniwn from /* to .4./-!.
To prove that PR is not ± Ali .
Proof. ProdneePQ to P' making Q/'' = i*Q. Ih-a.w RP'.
Then RQ 1 PP'. Tiyp.
P'Q^PQ. Constr.
.-. ieP = RJ^ and Z PiiQ = Z P'ii^g, Art. 79.
(*/. /'"o»i a ;w/h( t» a ± (0 a ghea line, tvto ohligne lines be drawn
cattiiiij o^ on the gii\n line eqiiiil segments from XMftiot of the X,,
ike oliliqae lines are eqaal anil make equal A with the L ) .
Bnt Pif^F is not a straight line, Art. G6.
(only one slniiijht line can be dnuvH eiiiiuecUng tiro gii'en piiiiiln),
:. PRI" is not a stmiglit Z .
.-. Z PRQ, the half of Z PRP, is not a rijjlit /. . Ax. 10.
.-. PR is not ± AP.
.'. onlv one perpendicular can be drawn from P to AP,
Q. E. ».
Ex. Tlll'BB
traiRlit liii
a iiiterse
t tit a pi)
aiiglos lormod
It tlie iKtiii
Me 30°
XQd 4U".
ftt the point.
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I'LANE (iKUMF/l'RV
TRIANGLES
81. A triangle is a portion of a plane homideil 1
straight liiiow, as the ti-iaiigle AUG.
82. The sidM of a triangle are
the lines which bound it; tlie
perimeter of a triangle is the sum
of the sides; the angles of a tri-
angle are the angles formed liy
the sides, as the angles ^1, B
and G; the vertices of a triaui>:le
angles of the triangle.
83. An exterior angle of a triaiigb
one side and by another side
produced, as the angle BCD.
With reference to the
BCD, the angles A and B
are termed the opposite inte-
rior angles.
84. Classiiication of triangles according to relative length
of the aides, A scalene triangle is a triangle in whieli no
two sides are equal. An isosceles triangle is one in wlueh
two sides are equal. An equilateral triangle is one in which
alt three sides are equal.
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TEIANGLES
33
85. Classification of triangles with reference to character
of their angles, A right triangle is a triangle one of whose
angles is a right angle. An obtuse, triangle is a triangle
(me of whose angles is an obtuse angle. An acute triangle
is a triangle all of whose angles are acnte angles. An equi-
angular triangle is one in which all the angles are equal.
86. The base of a triangle is the side upon which the
triangle is supposed to stand, as AB. The angle opposite
the base is called the vertex angle, its
angle ACB\ the vertex of a, triangle is
the vertex of tlie vertex angle of the tri-
angle.
The altitude of a triangle is die
perpend iciil.'ir from the vortex to the ^
base or base extended, as CD. '-'
87. In an isosceles triangle, the legs are the equal
siilys, and the base is the remaining side.
88. In a right triangle, the hypotenuse is the side oppo-
site the right angle, and the legs are the sides adjaeent to
the right angle.
89. Altitudes, bisectors, medians. In any triangle, any
side may be taken as the base; hence the altitudes of a
triangle are the three perpendiculars drawn cue from each
vertex to the side opposite.
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-1-i BOOK I. I'LANR fiEOMl?T!!Y
A bisector of an angle of a triaiiglo is a line wliicli ilividcv
tliis angle into two equal pavte. ThJH Liswtur is usiiiLlly pru-
diieed to mout tlto side opposite tlio givwi aiiglu.
A median of a triangle is a line drawn from a vertex of
the triangle to the middle point of the opposite side. How
many medians has a triangle ?
90. Two mutually equiangular triangles are triringles
having their corresponding angles equal.
91. Homologous angles of two mntnally equiangular
triangles are corresponding angles in thnwe triangles.
Homologous sides of two mutually equiangular triangJes
are sides opposite homologous angles in those triangles.
We shall now proceed to determine first, the properties
of a single triangle, as far as possible, then those of two
triangles.
92. Property of a triangle immediately inferred. The
SUM of any two sides of a triangle is giratfir than ihf third
mJp. For a straight line is the siiortest line between two
points (Art. 15.)
Ex. 1, Point out the hvpotiiesis and eonelusion in the generd onun-
tiation of Prop, 111 ; also point tLtm out in the pnrtienirLi' enunciation.
(As each of the uoxt Cftuen Props, is studivJ, let the pupil do tlia
same for it.)
Ex. 2. Find tlie angle whose complement is IS"; whose supple-
Ei. 3. It the eomplement of an angle is known, what is the
shortest way of finding the supplement of the auglef If the supple-
ment is known, what is the shortest way of finding the complement ?
Ex. 4. In 25 minutes, how many degrees does the minute-hand
of a clock travel 1 'Rtm m.iuy docs the honr-hand f
Ex. 5. Draw three straight lines ao that they shall i
three points; in two points; in one point.
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TEUNGLES SJ
Proposition IV. Theorem
93. Aiiif aide of a triangle is greater than the difference
between the other two sides.
Given A B any side of the A -4 liC, iiiul A € > iiC.
To prove AB>Aa—nC.
Proof. AB + BOAC, Art, 93.
(the sum ofawj two sitlcs of a triangle ?« i/rcaler lha» the thinl side).
Subtracting BG from eacli member of tlie itiequalitj-,
AB>A€—BC, A^.!).
'Sf equals he subtracted from unequals, the remainilers are uneijual in the
same order). q, s, D.
94. Cor. TJie perpendicular is the shortest line thai
can be drawn from a given point to a given line.
For, in the Fig. pnge 31,
PP'<FB + RI", Art92.
Or, 2 1'Q<2rR. Ax. 8.
.-. FQ<PR. A>^.V).
Hence, Dep. Tlie distance from a point to a line h the
perpendicular drawn from tjie point to tlio line.
Ex. J, If one sido of an equilateral triniifjle U 4 iiioliea, n-iiat i^ ita
perimeter ?
Es. 2, la it possible io form a triangle whose sides are 6, £1 and
17 inehea f Tt7 to do this witk the compaaaea and ruler.
Ex. 3. Is it possible to form a triangle in whii/ii one side is 10
inches and tho diiferein'o of tLe other two sides is I'l Inches f
Ex. 4. On a given line ns base, liy e.xm^t uae of ruler and uoni-
pftsaos, eonstrucL au equilateral tviauglo.
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VLAXE GEOAn'TltV
PltDI'OSITlON V. TlIKOliKM
95. If, from a point inth'm a irimu/U, i '■'■■ an ' ''■ nri'm
fo the extrfmilies of otie side of the triangle, Ike ^'ihi of
ihe other tiro xides of the trimigle is greater than the sum
of the two Uii';:^ no drawn.
Given P nny
FA and I'G lines
side AG.
lit within tho tvi!\iislfi ABO, aw\
WW from r to the extremities of the
To prove AB + BC> AF + PC.
Proof. ProaiicR the line AF to meet EC at Q.
Tlien AB + BQ> AP+ FQ. A
(a siraujIU hue !.s tlie siiortei line coiuieclMg tuo poiul^)
I Also FQ + QG> PC,
(s,i
n>).
Adding tliewe inecinnHties,
AB + BQ + PQ + QC > AP+ FQ + PC.
Substituting BC for its equal BQ + QG,
AB+ BC+PQ> AP+ FQ + PG.
Bubtraetiiig PQ from each side,
AB + BC> AP + PC.
Ex. Ozi a given line as base, by exact use of ruler and c
onBtvutit au ia03cel«H triangle each of whose lega ia double
. E. D.
irapassea,
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TKIANGLES di
Proposition VI. Theorem
96. Two triangles are equal if two sides and the in-
eluded angle of one are equal, respecthebj, to two sides and
the included amjle of the other.
id />/;/'' in which AB = I)E,
Given tlie triangles ^l HU a
AC^BF, and lA^ IP.
To prove A ABC = A DEF.
Proof. Place the A AB(7iipon the A DEF so that the
liue AG iioncide.s with its equal DF. Geoui. Ax. L'.
Then the line AB will take the direction of DE,
if,.rlA^ZDh!i h,j,>.).
Also the point B will fall on E,
(for line AE = line DE hj hyp.).
Hencfi the line BO will coincide with the line EF, Art. 6G.
{i^tlij one stiaiglil line oaH he drawn coimecting two yuiiU:,).
:. A ABC and DEF coincide.
.-. A ABC = A DEF, Art. 47.
(geometric figures lirRicft coincide are equal) .
q. E. D.
El. 1. What kind of pvoof is used in Prop. VI t (See Art. 02).
Ex. 2. IE S A, Baud C=CO°, 70°, 50°, J/J=10, Ji7=19, BC=iS:
also Z 7>=00°, I}E=\6, DF^IO: find -i E and F and side EF
wit bout measuring tbem.
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liOOa I. I'l.A
VnoroHvnas VII. TiiEOiiKM
97. Ttco ifUmi/l-'s are e-iiml if two anijli'^ 'ind fhr h,.
eluded sid<! of one ay cquiil, respet-ticdy, to iim angles and
the included side of the other.
Given Llic A ABC iiiid DEF in which ^A=ZD,
IC= IF, i\uAAG=llF.
To prove AAnC=ADEF.
Proof. Plaee the A ABG xipon the A I>EF so tliiit AC
shall coincide with its equal DF. Gsom. As. 2.
Theu AB will take the direction of DF.
(for lA^lDhyluji,.),
and the point B will fall somewhere on the line 1>E or DF
produced.
Also the line CB will tal^e the direction of FF,
{for LC = lFhjliyp.),
and the point B will fall on FE or FF prodnced.
.". point B falls on point F, Art. f>4.
{two straiglU hues ciin intersect in but one jioiixi].
.'. &. ABG and DEF coincide.
.-. AABC^ADEF, Art, 47.
{geometric figures toliich coincide are equoX). n e, B
Ex. 1. What kind o( proof is used in Prop. VII. !
Ex.2. If A A, B, 0=65°, 55°, GO', AB=2i. Aa=-['i, BC-2T:
alio 4 D, f=C5°, 60°, and J)f=18; find DE, EF, and I E,
Ex. 3. Conatriict by exiat iisa of ruler and compasses a scalene
triaujjle whose sides are 2, 3 and 4 times a givun line.
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TMANGLES Cif
Proposition VIII. Theorem
98. Two right iriangUs are equal if the hijpotenuse and
an acute angle of one are equal to the hypotenuse and ait,
acute angle of the other.
Given the riglit & A BC aiul DEI'^ in wliieii liypote-
nuse .4.iJ = liypotoimse DE, iiud lA = Z 1).
To prove
AABG=£X TiEF.
Proof. Pkfie tlie A ABG npon tiie A I>EF so tliat the
side AB shall coincide with its equal, the side DE, the
point A coinciding with the point D.
Then the line AC will take the diret^tion of DF,
{for lA^ZDbuhnp.)
Also the side BO will coincide with the side EF, Art. RO.
{from a given point, E, viithont a straight Itne, DF, but one ± ean be
(Irawn to the line).
:. A ABC and DBF coincide.
:.AABC=A.DEF, Art. 47.
{geometric figures ■which cuinckie are eqnal).
n equilateral trmngle, n
\. E, D.
li of whose sidea
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HOOK I. I'LANE GKOltETRY
Proposition IX. Theorem
99. Ill on isosceles trinn'jle the angles opposite- the equal
sides are equal.
Given the isosceles A ABC in which AB=^BC.
To prove ZA = /.C.
Proof. Let BD be drawn so as to bisect /.ABC.
Then, in the & ABD and DBG,
AB^BC. Hyp.
Also BD=BI), Ide»t.
And IABD= IGBD. Constr.
.-. A ABD- A CBB, Art, 96.
wo & are eqaiil if two sides and the included Z of one are egual, respee-
Uwly, to two sides and the included Z of the other. ]
.: /LA=lC,
{homologous A of equal A).
Ex. 1. On a given line as base, conBtruot exactly an equilateral
triangle above the line and another below it.
Ex, 2. On a given line as base, eonstruet esactly an isosceles
triangle whose leg shall be equal to a given line; make tlie same con-
Btrnetion below the given line and join the vertices o£ the two
lEOaceles triangles.
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TEIANCLER 41
Proposition X. Theorem (Converse of Prop. IX)
100. // two ait(]leK of a triangle arc equal, Iha xidei
opposite are equal, and the triangle is isosceles.
Given Ap frlanfflc ABC in which ^A=-- Z BCA
To prove AB = BC.
Proof. U thu sides AB and BC are not equal,
them must be longer than the otlier. Let AB be
thac BC.
On ^Sniark off ,10-BO Dmw DO.
Then, in the A ABG and ATJC,
AD^BC,
AC=AG,
IBAG^ IBCA.
((wo Si. are equal if iico siijes and the included Z of:
reapcctircly, to tico sides and the included L of the otlier) .
Ora part is equal to the whole, which is impossible. As. 7,
Hence AB cannot be greater than BC.
In like manner it may be shown that AB is not less
than BC.
Hence AB=BC.
Art. 90.
What ineti.od of lirooE
It ill a tri!iiigle VEF,
Dtaw u figure nui oil it 1
i usi-d hi Prop. X f
ai'fc the value o£ tbe I'ai
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BOOK I. PLASE GEOHlETIiV
Proposition XL Theorem
101. Tim triangles are cqmil if the thrrr .w'.'.s- "f mis
•e equal, re^pi'i'lirchj, to the- three sides nf tin- <)(h:r.
Given the A ABC iiiul BEF in wliieli AIi = l)E, I1C=
J3F, fnidAG^nF.
To prove A /U.'C= A BEF.
Proof. Place the A ABC so that its longest siile AC
shall eoineidu with its equal DF in the A DEF and the
vertex B shall fall ou the opposite side of DF from E.
Geora. As. 2.
Draw the line EB.
Then BE^VB (Hy-p.) .: A DEB is isosceles. Def.
.-. Ip^ /: r. Art. !)9.
(in an isosceles A Wie A opposite the equal sidun are equal).
In like manner, in the A BEF, tq=- Ls.
Adding,
L'P^ Lq= Zr+ Is, Ax. a.
Or II)EF=^ Z DBF. Ax. ii.
.-. A DEF = A DBF, Art. 9G.
{tKo A arc equal if Iwo Hides ami the iHi-hutei Z of one are c<iiial,
respcctirclij, to Iwo sides ami tlie inrbuled I of Die other).
:. A ABC = A BEF. Ax. 1.
— . — Q. E. D,
Ex. 1. Construct two equilateral triangles on the
same base, one abore and the other helow, and join the
two vertices. Prove that the line joining the vertices
bisects t1ifi Tf-rie^ iingl"a, and also bisects the base at
right angles.
Ex. 2. Hence, at any point in a given straight line,
construct exactly hy use of ruler and oompasseB a
perpendicular to tliat line.
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TRIANGLES 43
Proposition XII. Theorem
102. Two right triangles are equal if the hypotrniise and
a leg of one are equal to the hypotenuse and a leg of the other.
Given two right A ABC and Dl^F l!avm<; the hypote-
nuse AB =h3-poteimse DE, and BC=J-iF.
To prove A AB€= A BEF.
Proof. Place the A ABC m that BG shall coincide -with
its equal, EF, and A fall on the opposite side of EE from
D, at A' . Geom. As, 2.
Then A'F anA. ED will form a straight line, A'FI>, Art. 74.
(i/ tmo adj. A are toqetticr equal to two rl. A , their ej:t. sides form one
and the same straight line).
A'E^EIK Hyp.
,■, AA'ED is isosceles. Dt.>f.
;. /1A'= Z.D, Art. 99.
^oscelcs A the A opposite the equal sides are efiiiaL)
:. A A'EF= A DEE, Art. 98.
are eqiialif the hypotenuse and an amie i. of oi
the hypoteniwe and an acute I of the other).
AABG=ADEF. a^-
__-__^_^~_ '■ ^- "•
But
(i«
ray given stcaigLt
Ex. 2. Biaect any yivBii imgk AOB,
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BOOK I. PLANE GEOMETRY
rROl'OKITlOX XIII. ThEOEEM
103. All erierior iniijJe of a iriangk is greater than
thcr iippiixite interior anf/k.
Given / BGI) an esterior Z. of the A ABC.
To prove Z SOD greater than lAJWor /BAG.
Proof. Let E be the middle point of the line IIC,
Briiw AE and produce it to F, making FE^AE. Draw
FC.
Then, in tlie A AEB and EEC,
AE^FE, and BE - CE, Constr.
I BE A= Z. FE€{Mng vertical i). Art. 78.
.-. A ^ J:B - A FEC, . Art. %.
(frco & are (qual if lico aiih-s and the indiiilril I of one are egtial, re-
spet-lk-ciy, to two iiik-s and the indmlcd I of the other).
:. ZABE= IFCE,
{being bomologmis d of eiiiial A].
But Z BCD is greater than Z FCE, Ax. 7.
{the ic/iD?e is greater than any of its parts).
Substituting I ABE for its equal ZEOE, Ax. 8.
ZBCB is greater than I ABE, that is, than lABG.
Similarly, by drawing a line from B through the midpoint
of jj (7 :ind by producing SC through (? to a point .ff, it may
be shown that lACH {= IBCD) is greater than ABAC.
Q- E. D.
eacU of wLose legs eqaala balf a given line.
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TRIANGLES 45
PROPOsiTioN XIV. Theorem
104. If two sides of a triangle are unequal, the angles
opposite are unequal, and the grette-" angle is opposite the
greater side
Given the sida BC > side AB in the A ABO.
To prove Z BA C greater than Z ('.
Then, in the isosceles AABD, Z)'= Zs, Art. 99,
{in an isosceles A the £s opposite the equal sides are equal).
ZjB^(7 is greater than Zr, As. 7.
[the Kliole is greater than anij of its parts).
.■. ZJBAC is greater than Zs. Ax. K.
But Z.s is ,111 exterior Z of the A ABC.
:. Z/{ is Ki'eater than Z C, Ait. loa.
(anext. Z of a A is greater llnm eUltor opposHe iiil. Z ].
Much more, then, is Z BA€ {which is greiiler than Zs)
greater than Z C, Ax. 12.
(if, iif thrre (fiinniities, the first is greater Ihnn the secmid, OHii the Keeoiiil
is greater than the third, then the first is grenler than the Ihu-il) .
Q. £. D.
105. Note. The esaeutial afeps of the above proof may be
arranged in a single statement, tUua:
lJ<AlJ>Zr = ls>lV .: IBAC is greiiter than IC.
Ex. J. Which is the longest side o£ a riglit triangle ? of an obtuse
triangle f
Ek. 2. Conslriiot exactly an equilateral triaugle, Bttth of ivhoee
Bides is half a ^iveu line.
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PLANE CEOME'l'HY
PKOi'osiTiON XV. Theorem (Converse op Peop. XIV)
106. If two angles of a triangle are unequal, tie sides
opposite are luieqiial, and the greater side is opposite the
greater angle.
Given I A greiitei- th;iii Z C in the A ABC.
To prove BC > AJi.
Proof. BC either equals AB, nv is less than .4 B, or
greater than AB.
Biit BC cannot eqna! AB,
for, if it did, /.A would equal Z C, Art,'
(being opposite equal sides in an isoscvles A),
But this is contrary to the hypothesis.
Also BC cannot be leas than AB,
for, if it were, Z A would be less than Z <7, Art. i
(i/ two sides 0/ n A aye unequal, the i amiosite are uueqattl, ami
greats Z ia opposite the tfreater side).
This is also contrary to the hypothesis.
;. BOAB,
(fzr it ncilhcr equals All, uor in !ras than AH).
Q. E. B.
El. I. Draw a triangle the altitude of whieh fuUa on the b
produced. What kiud of a triangle is tliis i
Ex. 2. Draw a triangle the altitude of
which coincides with one side. What kiud of
a triangle is this f „^ — .^^
Ex. 3. By exact use of the ruler and com-
passes, dww a perpendicular to 6 given liue
from a given point without the line, ' "
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TMANGLErt
PftorosiTlON XVI. T1IEORH.M
107. If two triangles have two sides of one equal, respec-
tively, to two sides of the other, but the included angle, of
the first greater than the included angle of the second, then
the third side of the first is greater than the third side of
the second.
Given tli(! A AIW and J)EF in ivincli AI1 = DE,
BC=EF, and lAtW is greater than I.E.
To prove AC > liF.
Proof, Place the A I)EF so that the side DB coincides
with its equal, the side AB, and I*' takes the position F' .
Geom, As. 2.
Let the line BE bisect the IF'BG and meet the line
AC at H. Draw F'H.
Then, in the A F-BH and BBC, F'B^BC. Hyp.
BH^ BE, Went.
IF'BE^ ICBH. ConstT.
.-. A F'BH'^ A BEC. (Wby?)
.', F'E=CH, {homolot/ous sides 0/ equal ^).
Hut AR + EF' > AF", Art. 92.
{Die Slim oro'll ''f skies 0/ a A is greater than the third shU],
SubslitutiiiK for EF' its equal EC,
AE + EC, or AC> AF'. Ax. a.
:. AC > 1)F. A.,. 8.
q. E. ».
Ex. 1. Draw a figure tor Prop, XVI in which the sides and au-
gles are of sneh a size that i'" falls within tlie trianftle J fir.
Ex. 2. Uraw another figure in whicli *" fiUU on the side AC.
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43 llOOK I. PLANE GEOMETRY
ritOfORITlON XVII. ThBORE.1[ (CONVERSE OF PUOI'. XVI)
108. // tico sides of a irimigh are equal, respectivehj,
to two aides of nnotJier trimigle, but the third side of the.
fii-xt is greater than the third side of the second, then the
unfile opposite the. third side of the first triangle is greater
titan the angle opposite the third side of the second.
Given thp A ABC iind HJ^F hiwing AB^-DE. B0=
EF, but AC > DF.
To prove ZJJ greater tliuii ^E.
Proof. The Z.B either equals Z.E, or is less thunZE,
or is greater than Z E.
But Z B does not equal ZB,
for, if it did, A ABC -would = A DEF, Art. or,.
{tiro A are eqiial if tico "iilcs and the ineluileil I of one are equid,
rcspeclh-etij, to tiro sides ami the iiieluihil Z of the other),
and. AC would eqiiiil DF (Jtomoiogoua sides of equal &^) ,
which is contrary to the hypothesis.
Also if AB were less than Z£,
side AG would be less than side DF, Art. 107.
(if two A liavetwosides of one equal, respeelivelji, io two sides of the olhtr,
hut the itictaded Z of the first greater tliait the iacludeil L of
the second, then the tAird side of the first is greater
than the third side of the second).
But this is also contrary to the hypothesis.
Hence ZB is greater than ZE,
{for it neither equals I E, nor is less than ZE).
Q. E. 9.
Ei, On a f-iven line (0 con-
nictt a triangle whose other two siiles
■e equal to two giyeu lines {m and »).
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LINES 49
PfiOPERTIES OF tlHES PROVED BY *ISE OF TSIAWGIES
PRorosiTioN XVIII. Theorem
109. Of lines drawn from the same point in a perpen-
dicular and euttinij off unequal segments from the foot of
the perpendicular, the more remote is the greater.
Given PO 1 AB, FT and PQ oblique to AB, aud
OT > OQ.
To prove FT > PQ.
Proof. J'roani^e PO to the point P' making OP'=OP.
Oil AB take OR=OQ. Draw PE, P'Q. P'R, P'T.
Then I'Q = PR, Art. 79.
{if^froin a pnint in n ± , a gii-cit line, two oblique linen be drawn, mitliiig
off oil the gieai line ri/iiat segmtnts from the foot of the X , the
oblique lilies are equal).
Ill A PTI", PT+ TP'>PB-VRP', Art. 93.
[if, from a point u-illiin n A, two lines be drmnt to the extremilies of a
side of the A, the sum of the other two sides of the A is greater
than the sum of the tm lines so drawn).
But oris X PP and PT Mid P'T cat oft equal segments,
PO and i:"0, fi-crni the foot of the L AO.
Hence PT^ P'T. In like" manner PR = P'R. Art, 7£».
Hence, by substitution, 2 PT > 2 PR. Ax. 8.
.-. PT > PR. Ax. 10.
.-. PT > PQ. Ax. a.
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PLANE GEOMETRY
Proposition XIX. Theorem (Converse of Prop. II)
110. Equal ohlique lines drawn from apoiiU in a per-
pendicular cut off equal segments from the fool of ike
perpendicular.
Given PC ± AB, PK and FT oblique to AB, and
PR = FT.
To prove CE=Cr.
Proof. Ill the right A RFC aiul CPT,
PG^PG. Ident.
Also PE^PT. Hy,.
.-. ARPG^ACI'T, Art. ir-
{tico right iJi are equal if ihehijpoleimse ami a leg of one are equal !•• ir,.
hypotetiuae and a leg of the other).
:. RG=CT,
{liovtologotis H'ulei) of eipial ^).
1). E. B.
111. Cor. Of two unequal lines draion from a poini
in a perpendicular, the greater line cuts off the greater
segment from the foot of the perpendicular.
Thus, if PT> PQ (Fig. of Prop. XVIII), OT cannot
= OQ (Art. 110) ; nor is OT < OQ (Art. 109) .-. OT > OQ.
Hence, also, from a given point only two equal straight
lines can be drawn to a given line.
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LINES 51
Proposition XX. Theorem .
112. I. Efery point in the perpendicular bisector of a
line is equally distant from Ihe extremities of the line; and
II. Eeery point not in the perpendicular bisector is
unequally distant from the extremities of ihe Una,
Given the middle point of the line AB, OC ± A P..
P any point in 00, and Q any point not in OC.
To prove AP^^PB, bnt QA and QB unequal.
Proof. I, AP^PB, Art. 79.
(if, from a point in n A- to a gii:en line, tifo ohliqiie lines he drtiwit
cutting off OK the given line equal cutji'iinttn, eb:.).
II. Since Q is not in the line OG, either AQ or QB must
cut the line OG.
Let AQ intereeet OG in the point R and join RB.
Then AR — RB, [bij first pari of this theorem).
To each of these equals add RQ.
Then AR + RQ=RB + RQ. Ai. 2.
But RB + RQ > qP. .Wiiy?)
.-. \>v substitution, AR + RQ, or AQ > QB. Ax. 8.
Q. E. D.
113. Cor. Two points each equidistant from the ex-
tremities of a line detertnine the perpendicular bisector of
ihe line.
This corollary gives a useful method of determining tlie
perpendicular bisector of a given straight line, by determin-
ing two points only of the perpendicular bisector.
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I'lAKK LiEOMF/niY
114. Dep. Tlio locus of a point is the pnlli of a point
moviuL' a<!i;or(lin^ to ii given giiometrii; hiw.
Tims, if a point more in a plane so as to be always two iiiehes Jia-
tfliit from a given point, its tocua is tlie cirenraEereiiue of a circle
wliose center is the given point, and wliose radius is & line two iuPhea
in length.
TliuB, also, the locus of a point moving so as to bo eqiiidisf!i,nt
from two giren parallel liuos is a strait'lit line lyiug midway butween
the two given lines.
The locus of a point mny p,onKist of two or more sepa-
rate lines or parts.
Thus, the locus of a point moving so as to be always at a given
distance from a f;iven line is two lines, one on either side of tlia
given line, at the given distance from it.
115. Demonstration of loci. In order to prove tlint a
given line is the locns of a given point moving according
to a given geometric law, it is necessary:
1. To prove iJiai every 2>oiiit in the (jireii Hue i^irfisflrs tlie
given hnr or condilion.
2. To prox'c that every jwini not in the {/irrn line does not
sdiisfif Ihe given law or comHtion.
Instead ot 2, it may be proved tiiat every point which sjilisCes t.he
given condition lies in the given line.
Henee, in Prop, XX it has been proved that the per-
pendicular bisector of a line is the locus of all points equi-
distant from Die extremities of the line.
116. Use of loci. Loci are useful in determining a
point (or points) wLieh shall satisfy two or more geomet-
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LOCI 53
rieal conditions. For, by finding the loens of all points
which satisfy one of^the given conditions, and also finding
the locns of all points which satisfy a second condition,
and then finding the intersection of these two loci, we
obtain the point (or points) which satisfy both coaditious
at the same time
Thus, if it be required to find the poiuts wliich are two inubes
from one glvon poiat and tbrne inchea from another given point, tbe
two given points being four inubos apart, the required points are tbe
intetsections of tbe circumferences of two t-irtles.
Let the pupil inabo a construction and obtu,ia tho riiquiroJ point.H.
Ei. 1. Dran
inch from a piv
f the locUB of rt point movi
en point.
iiK at the distal
ice of
one
Ek. 2. Drav
one inch from a
7 exactly tbo locus of a poii
, glvon line.
at moving at a distanci
9 of
Ex. 3. Driv«
distant from the
r exactly the locus of a poii
1 extremities of a givou line
1 one iuohlong.
..= .„,.
Ex 4 n w
tworaralicl 1 n
man^ p i ts 1 1 -x plane ,
OS ' Ihieo panllel hneb !
(■^te irt 4- )
tule 1
Ex 5 \r
porrespouJ u„
two t 11 les Ljiil t th
Ihr L ingk-* 1 tile tl r
e ingl ^ of one
t lHufttr.ite by
1. al
d iwii
the
ij, a
Ex 6 1 a
conne t the i ^
V th ee IS epbIob tiian ks on the same
irti es Wb it truth is ilIustratLd 1 y th 3
ijabe
figure
„d
Ex 7 Draw a strait-ht 1 no ind locate a po i t " inches tii
By the use of loei locate the po nts wh ch aie 1^^. incbes fi)
given line iind aX the same diEtau>,o from the i, ven point
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54 BOOK I. I'LANE GEOMETIIY
Proposition XXI. Throrkm
117. I. Evpry point in the iiseclor of ini angle
diatani from the skies of the angle; aud
II. Co^VERBViLV, every point equidixiatit from i
of an angle lies in the bisector of the angle.
I. Given TB the bisector of the iuieie ABC. P i
poii5t hi PB, I'ij J. /:.L, iiiui FB ± no.
PQ^FK.
lu the rt. A PBQ aad PBB,
PB^PB. Id
ZPBQ=ZPBE. 1
:. APBQ^APBR, Art,
■c equal if the hypoteimse aii<l ait
To prove
Proof.
Also
{tu
'- ofo
Hence
[honiolugotis siJes of equal &).
II, Given ZABC, PQ1.AB. PE±BG, and PQ = Pli.
To prove tlisit PB is the bisector of /.ABC.
Proof. In the right & PBQ and PBS,
PB^PB. Ident.
Also PQ=PB. Hyp.
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Or
.-. A PBQ^-APBE, Art. 102.
re equal if the hypolouise and a Icfj of onf, sU:. ) .
:. Z AiiP= Z GBP (lio-mhgous A of equal A).
ZABCh bisected bv BP.
Q. E. D.
118. Cor. In Prop. XSI it has been proved that the
bisector of hh ntigk is the hens of all pohils fiquMistanl from
the sides of the awjlr, for it h;i5 beon proved that every
point in the given Hue siitisfioa tlio given law or condition,
and that every point which satisfies thti given condition
lies ill the giveLi line (soe Art. 115).
E
119. Dr.p. A transversal is a
line that intersects two or more
other lines. Thus, EF is a trans-
versal of the I'nes AB and CD.
If two lines are cut by a trans-
versal, it is convenient to give the
eight angles of intersection special
names.
n, h, 0, h, are called exterior ;uigles.
c, d, e, /, are called interior angles,
c, /, form a pair of alternate-interior angles.
b, /, form a pair of exterior-interior angles on the
same side of the transversal.
Let the pupil name another pair of alternate-interior
angles; also name another pair of exterior- interior
angles on the same side of the transversal; also name a
pair of interior angles on the same side of the trans-
veraai.
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56 iiOOK !. PLANE GKOMETRY
PARALLEL LIKES
120. Def. Parallel lines have already iieen ileiined
(Art. 4:1) as stmiglit lines wliich lie ic the wimo piano iiiiil do
not meet, howcvci' far thoy bu produceil.
What is the fuiic!ameutal axiom eonccrniiig pafallel
lines? (see Art. 47.)
PROrOSITION XXII. TUF-ORKM
121. Two straight lines in Ihesatne plane, perpendicular
to the same straight iiite, are parallel.
Given the lines AB and CD J. line FQ i\iid in the
same plane.
To prove AF, || CD.
Proof. If AB and CD are oot parallel, tliey will meet
if sufficiently prodiieod. Art. 120.
We shall then have two Js frura the same point to the
line PQ.
But this is ii
Heace AB and CD never meet.
,■, AB and GD are parallel.
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PAltALLEL LINES
57
122. Cob. Twosiraight lines parallel to a third straight
line are parallel to each other;
JAnes parallel to parallel lines are parallel;
Lines perpendicular to parallel lines are parallel;
Lines perpendicular to noii-paruUel lines are not 2>arallet.
Proposition XXIII. Theorkm
123. If a alraii/M line is perpendicular lo mir of iu:o
given paraUcl lines it is perpendicular to the other also.
Given An II CD, i\w\ PQ L AB.
To prove PQ 1 CD.
Proof. Let or be drawn X FQ at 0.
Then CFWAB. Art. m.
IfKo siraigltl lines in the savic plane X snmr. sirnUjhl liiw are ||).
But
{through n. giccn i<oi
CD II AB.
CF coincides with CD,
I one straight- line, aiul only i
aiioOicr gir':a straight iinv).
Bnt
Hence
Hyp.
{for CI) coincides with CF, to ii-hich FQ is X)-
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DO BOOK !. rLANV; Ci;(>?ilETItY
Proposition XXIV. Tukokiiji
124. // I'ro parallel sirnigM Uihx irrr ■■nt hij a trans-
tersal, llw allcruate inlcrwr <iiighs '.irc equal.
Given the 11 lines AB and C7' (;ut by the tt-aiisvevKai
I'Q at the points G and E respectively.
To prove lAGE^ KjED.
Proof. Thi-ongli R, the middle point of EG, let the line
ffFbe drawn J. AB.
Then HF X 6'/), Art. 1^3.
Ufa straight line is X one of tico i] (iiiw, il is X the oiiur alsii).
, 111 the right A QJtH and EEF,
GE^EB, Consw.
/tOMS^-ZEBF. (Whjl)
.-. A (Jffiff = A ERF, Art. 98.
(too right A (wa ejwa^ ^ the hypotenuse and an aauie Z of one =:
(Ae hffpot. and an acute / of Vie other).
:. I HGR = Z REF, or, IA0E= I OED,
Uiomologous A of equal A).
Q. E. B.
Es. In tte above figure let the pupi! show that Z BGE = Z GBC.
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PARAiLEL LINES O'J
Prop. XXV. Theorem (Conveese op Prop. XXIV)
125. If two simujhi lines are cut hij a transverml,
making the aUenuite interior angles equal, the two straight
lines are parallel.
^Given tlie t'wo lines AB and CD cut by tlie triinsveraal
PQ at the points F and ff, making Z AFG - I FGI>.
To prove AB \\ CI).
Proof. Throngh F let the line KL be dmvvn li CI).
Then
{iftivo II si.
Bnt
. lines
I KFG = Z FGl),
are ciU hy a transversal, the all. int. S a
IAF6 = IFGD.
.\i-t. 124.
rv equal).
Hyp.
Hence
IKFQ=- I AFG.
:. KL coincides with AB.
As. 1.
Bnt
Heuce
(fo>
KL II CD.
AB 11 CD,
Al! •:oincides milt KL, wlurh is \[ CD).
Conatr.
Ex. 1 . Tf Z BFG = Z FG C, prove that AB aid CD a
Ex, 2. By exact
use o£ rulai' and
eompasaes , lit a
given point {P) iu
agivenstraigbtline
{OA) poivstiiiot an angle equal to a given Z(B),
Ex. ;
lelti
lethods, tbrougti a giv
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PitOPOSiTiox XXTI. Theoieem
126. 7/ ta-oparnlM lint^s are cut bij a transversal, the
exterior interior (Niy/i-s tire eqiuil.
Given the || lines AB untl CI) cut by Uie transvprsa! PQ
at the points Fand G respectivelj-.
To prove ZPFB^ZFOD.
Proof. Z FFB = Z AFG. (wiiy! )
IFGD= Z ,-lF(7, Art. 121.
Cbeiitg all, uii. A of inirnlld Hiics).
.-. IPFB^Z FGI>. As. 1.
lu like maniipi- it may be shown that AFFA - Z FGC.
q. E. B.
Prop. XXVIL Toeori^m (t'oxvEKSK ov Prop. XXVI)
127. If Uvo HtraUjlii linn arc ad J»j n irani:rn-Kal, nuik-
ing the exterior interior angles equal, the two siniiqhi lines
are parallel.
Given, on Fig. of Prop. XXV, ZPFB = IFGD.
To prove AB 11 CD.
Let the pupil supply the proof.
Ex. If, ill l^lie Fig. to Prop. XXVI, /. ['FB equaU €.7°, Ciad tho othi.1
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PAEALLEL LINES Gl
Proposition XXVIII. Theorem
128. If two parallel lines are cut hy a transversal, the
sum of the interior angles oii the same side of the trans-
versal is equal to two right angles.
Given the ll lines AB and GH ent by the tmnsversal PQ
at tlie points F and 6 respectively.
To prove Z BFO + IFGI)=2 vi. A .
Proof. /iFGI> = ^FFIi. Art. ISO.
To VM-], of tliese eciiials add Z BF(i.
Then / BFO + IFGD= IPFB + Z BFO. Ak. 2.
But Z FFB + Z JiFO^-1 rt. d . A,t, v.i.
.-. IBFG + -^FOB^^rt. A. Ax. i.
Q. £. D,
Pkoi>, XXIX. Thf.(irf.m (CotiVERSF; oe Prop, XXVITl)
129. // iiro straight lines are- nit hy a transversal,
maMng the stun of the interior angles on the same side of
the trai»iversal hiuuI to tieo right augles, the dm lines are
varalkl.
Given, on F\<r. of Prop, XXV, Z BFG + Z F0JJ = 2 rt. A .
To prove AB \\ CD.
Let the pupil supply the proof.
Ex. If, ou Fig. of Proi.- XXVIil, ^i'Fli + ^QGD^IBO", are Jli
«d CD II r
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? BOOK I. PLANE GEOMETKY
Proposition XXX. THrcoKicjr
130. Two angles whose sides are paralld, vach to each,
•e either equal or supplei.
Given A7i || JiR, and EC II GF.
To prove A.ABC, DBF and GKIT pqiial, and AABC.
and DEG supplementary.
Proof. Produce the lines JiC and Hl^ to interseet in P.
Then Z ABC = I QPE, and Z QFE = Z DEF, Art. 120.
(fceinp eri;i. int. A of || lines).
.-. IABG= IDEF. (Wliyf)
Also ZDEF= ZGER. CWliy!)
.-. lABC^ IGEH. (.WhyO
Ajjain .i Ti/JF and BEG are siipplementaiy. Art. 73.
.-. A ABC and DEfJ are snppleraentai'v. Ak. 8.
Q. E. B.
131. Note. It is to be observed tlmt in tbe above theorem the
two angles are equai if, \a the pairs o£ parallel sides, hoth pairs estend
in the same direotiou Irom tiie vertices ( i B and DEF), or both paira in
opposite directions ( i B aod GEH) i and that they are supplementary it
tmepair extends in the same direction, aud the other pair in opposite
directions { ii B and DEG). The directions ot the sides are deter-
mined by connecting tlie vertices of the angles and observing wliether
the liaes considered lie on the same side or on opposite sides of the
line drawn.
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PARALLEL LINES
PsoposmoN XXXI. Theorem
132. Two anglfK ;
each are either equal a
hose sides are perpendicular each to
supplementary.
Given BA ± EF, and nC X TiD'.
To prove ^J.J5C'= IFEI), ami A ABC and FED' sup-
plementary.
Proof. At E let line EK be drawn ± EF and in tlie
same direction with BA ; also EG L !>!>' and in the same
direction with BG.
Then BA 11 EK, and JJC 11 EG, AH. I2i.
(iji'o sirciiijht lilies in tlie same plane, J. (lie same straight line, are H),
.-. LABG = /LEEG, Arts, m, ni.
((wo ^ whose sides are ||, eacft to each, aiid extend in the same lUreelii/n
from the vertices are =).
But I KEG is eoraplement of IBEK, Art. .t^.
{for IDEGisart. I lif coiislr.) .
Also Z.FET> ia eoniplement of IDEK, Ait. 33.
{for I FEE is art. L by conslr.).
:. IFED^IKEQ, Art. T5.
{,eoT>ipleme»ta of the same L are = ).
.-. LABO=LFEJ). Ax. 3.
But I FED' is supplement of / FED. Avt, :i4.
.-. ZFi'i'' is aupplement of ZAiiC. Ax. 8.
Q. E. D,
133. Note. In the above tteorem the two angles ore equal if the
sides, oonsidert'd as rot.iting about tiio vertices, are taken in the same
order (thus SG ia to the right of BA, and ED to the right of A'."'
■'■Z.ABC= ^FEV); but tlie angles are Bupplementary if theeorre'i-
ponding Bides are tiik-flii in tlio opposite onltr (llins,^Cis to the rii:lit
ttSAXKHEiy is to tku left of EF :. Z JZ)'C' = Buppleineiit et IFti:!').
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PROrOSITIOX XXXII. TuEOTiKJI
134. The Slim, of the. unyles of a Irian.,},' h i'qmil io tiv
rUjhi migli's.
Given tlifi A A BC.
To prove lA+ZIi + l nCA - 2 vt. A .
Proof. Pi-odnfie the siile AC totbe point. 7) sniil theon^h
C let the line CE be dmwn || AB.
Then Z Kr7) + Z iSCB + Z .fiCA = 2 rt. ^ , Art. 77.
fi/ic siiiH i)/n!i llie A alHiiit a point on the same side of a su-tiiijlii line
pii.'isiiiij tliiojigh the poi>it = 2 rt. S. ).
But IBGB^LA, Art. 120.
{fe/Hfif (xi. ini. ti. ofpnruUel U»es).
Also IBCE = ZR, Art. V2i.
{hil,u, nli.int. A ofp'ir„!lfUiw,i).
tSiihslitiitiiigfi.r ZICCD itseiiual, /.4,ai)d fnr IBCE
it.^ equal, Zli, Ax. 8.
/A + zn + ZnVA^'l vt. A.
0- E. D.
135. Cor. 1. An exterior angle of a triungU is eqwil to
the sum of the fico opposite interior angles.
136. Cor. 2. The sum of any two angles of a triangle
is less than two right angles.
187. CoE. 3. In a right triangle the sum of the two
acute angles equals one ri'jht angle.
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PARALLEL LIN'ES 65
IS8. Cor. i. A triangle can have but one rtgM. or one
obtuse angle.
139. Cor. r>. If two angles of one triangle equal two
angles of anolher tnangJe, the third angle of the first tri-
angle equals the third angle of the second.
140. CoE. 6. If an acute angle of one riyht triangle
equals an acute angle of anotlmr right triangle, the rentain-
ing acute angles of the triangles are equal.
141. Cor. 7. Tiro triangles are equal if two angles and
a side of one are equal to two angles aiul the homologous side
of the second.
142. GoR. 8. Two right triangles are equal if a leg and
an acute angle of wie are equal to a leg and the Iwmologous
acute angle of the other.
Ex. 1. If two nngips of tt triiinelG sire M° ami G'i^", find the re-
maining angle.
Ex.2. If one nciito .nngle of n, riglil trinngle is r>fi° 1"/, find Ihe
other acute niiglc,
Ex. 3. How xnKny lifgi'Gos in each angle of an equilateval triangle?
Ex. 4. IIow nianv degrees in eatli acute angle of an isosceles
nsht liianslo!
62", 72°!
Ex. 6. If one angle of a triangle is 4^'°, find the sum or the olher
two angles.
Ex. 7. If two nngles of a triangle are 3S° and 65°. find all the
est«rior angles of the triangle.
Ex. 8. If tlu^ verti's aiifjie of an isoiicelea trinngls is 3S°, fiml eaeli
Ex.9, If an angle at the Use of an i>,llSl^eios triangle is .'iO", find
tbe vertes ani;li-,
Ex. 10. All f\terior aiisle at tlie liasu of an isosceles triangle is
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I'LANK <;i:oiiini(Y
QUADRILATERALS
143. A quadrilateral iri a portion of a piano boumlod.
by four straiglit lines.
The sides of a quadrilateral are the Ijounding lines; the
angles are the angles made by the bounding lines; the
vertices are the vertices of the angles of the quadrilateral.
The perimeter of a quadrilateral is the snm oi the sides.
144. A diagonal is a straight line joining two vertices
that are not ailjacent.
145. A trapezium is a quadrilaieral no lno o( wlioso
sides are parallel.
146. A trapezoid is a quadrilateral whicli lias two, and
only two, ot its sides [Kiiallcl.
147. A parallelogram i;
sides are parallel.
a quadrilateral whose opposite
14S. A rhomboid
oblique angles.
149. A rhombus is a rhomboid whose sides are equal.
150. A rectangle is a parallelogram whose angles are
right angles. g'
151. A square is a rectangle whose sides are equal.
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QUADEILATEE.U^S
152. The base of a parallelo- c
gram is tlie side upon which it is /
supposed to stand, as AB. The /
opposite side is called the upper "*
base (CD).
The altitude of a parallelogram is the perpendicular dia-
tance between the buses, as JHF.
153. The bases of a trapezoid are its two parallel sides.
The legs of a trapuzoid are the sides which are not parallel.
The altitude of a trapezoid is the perpendicular distance he-
tweeii the bases. The median o£ a trapezoid is the line join-
ing the midpoiuts of the leg:-.
154. An isosceles trapezoid is a trapezoid whose legs
Ex. 1. Draw a quiidriltiteval with three acute
Ex. 2. Is every rhombus a rhnuitioid J Is
thombus ?
Bx. 3. \Thatis the ilifferenee b(
What [iropeilies do they have
Ex. 4. Find the perimeter o£ a square foot
Bj Ibid of the following clflBSificatfoii:
glee and one ob-
^ery rhomboid a
iqwui'e and a vhombuef
Quadrilateral
^ Trape^^oid ,
I. Pavallelogr
. Isosceles trapez
) Rhomboid . . . rhombus.
Ex. B. Determine what four mimes the rhombus is entitled to.
Ex. 6. Datevmiiie what praporties Ihe rbombua, square and
rectangle have in commou,
Ex. 7. A diagoniil of a rhombus divides tha rhombus into how
JUany tviaiiglea F What kiud ot triangles are these {
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BOOK I. I'LASE (;i:OMETliY
Propositiom XSXIII. Theoreji
1 55. The opposite sidi-s of a parallelof/ram are equal, and
its opposite an'jtfs (ire also equal.
Given the paralklograni ABCI).
To prove AJ)=BC, AB^'DC, in=^D, aud lBAJ>-=
I BCD.
Proof. Draw the diagonal AC.
Then, in the A ABC and ADC>
AC^AG, (Why?)
IBCA = ZCAD, Art. ui.
{being alt. int. i of parallel lines).
ABAG= AACD,
(some reason).
:. £^ ABG = C^ AGB, Art. nr.
((H'O A are cqwil if ia-ii A and ihn iiielii/lcd Kide of one an equal
rc.iperliri'ly U> liio A aii'l the. incbidvl side of the other).
:. AD^BC. A «= BC and Z B= Z 1),
ih,moloyo„s pans of equal &).
Ill like maimer, hv drawing the diagonal BD, it may be
proved that IBAB^IBCD. ^ ^ ^
156. Cor. 1. A diagonal divides a parallelofjrain into
two equal triangles.
157. Cor. 2. ParallelUnes comprehended between par-
allel line.i are equal.
158. GoR. 3. Two parallel Uhes are everywhere equi-
distant. ^ ,
Ex. 1. In the above figure, prove ^SJO=ZflCD byuae of Ax. 2.
Ejc. 2. Prove tiie oppoeite angles of a, p stall el og ram equal, by use
0£ Art. 130.
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quadeilateeals; by
pROPOSmoN XXXIV. Theorem
159. If the opposite sides of a quadrilateral are equal,
thfi figure is a parallelogram.
Given the quiuli-il^t.eral ABCT) i
F.€=AT>.
To prove ABCl) a dl .
Proof. Dra-sv the diagonal AG.
Then, in the & ABC and ADC,
AC=AG.
(Why ?)
BC^AT).
(Why?)
AB=CD.
(Why 5)
.: AABG^AADC.
(Why ?)
.: Z E.-1C= Z AGT>.
(WhyT)
:. AB II CJ>,
• (ICO lines are ,:i<t h,/ i, Irami-crsal, miihii'i the alt. i
IhKl: arc |j).
Art. 125.
III. i. equal, the
Also Z£eA = I CAD.
(Why?)
:. BG II AD.
Art. 12,-K
.'. ABOI) is a C^,
a ijHUitrilaleral tcliusv ojipo.
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70 iXlOli; I. PLANE OEOlIETliY
PiiOPOsiTioN XXXV, Theorem
160. // tivo sides of a qiimh-ilateral are equal and par-
allel, the other tiro sides are equal andpuraUcl and the figure
is a paralletoyrain.
Given tlie qiiaaiitaterni A liCD in wliieli BG = and ); AIJ.
To prove AB(JD a d/ .
Proof. Draw the diagonal .4 C.
Then, in U
ic A ABC ami ADC,
AC=AC.
(Why!)
BC^AB.
(Wby 1)
IBCA= ICAT).
{bchig •ill. u,l. A of jMuilM Unr^)
Art. 124.
:. A ABC^AABC.
(Why?)
.: I BAC=IACB.
(Wliy t)
tifo Juics arc
:. AB\\ CB,
Ml by a lnii::<i-fT.i(\l, mildinj Ike all
Iwes are [[) .
. /(If. A
Art. 125.
(■.,"ll/, Ilw
.-. ABCI) \& 3. rj .
■nt augl.
Ai't. 14V
J. E. D.
Ex, 1. Shon
that in a Z35 each pair of adjai:e
plementary.
Ex. 2. One angle of a parallelogram is 43° ; find the ottier an
Ex. 3. If, in the triangle ABC, Z J=60°, Z2i = 70°, whicli i;
longBRt Bide in the triangle I Which the ehorteat !
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QUADRILATKllAI-S 71
Proposition XXSVI. Theorem
161. The diagonals of a parallelogram bisect each other.
Given the diagouuls AC aiul Bl> of the £UAHC'D,
intersecting at F.
To prove AF=FG, and r.F=FTi.
Proof. Let the pupil supply the proof,
[SuG. In the A BFC and AFD what si.les uro equal, aiul wby ?
What A are equal, and why 1 ett.]
£eure V
Ex. 2. If one aiiKie III n parallelogram i^ tlu-ee (iiiiL-saQotliev angle,
find ali the angles of l!ie parallelogram,
Ex. 3. If two angles of a triangle ace j." ami v°, find the third
angle.
Ex. 4. If two angles of a triangle are j" and OU^ + j°, find the
third angle.
Ex. 6. If OBB angle of a parallelogram is a°, Ihid thu other uncles.
Ex. 6. Construct exactly an angle of W.
Ex. 8. liow hirfjB way the doutile of an ubtuab uuyie be ! how
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l-i 1500K I. I'LAXi; GKOMiyrRV
Piim*osrrio.N' XXX VI i. TiiEdiii-.M
162. Two parnJMogramH ui-'' eqind if two ailjticcnt sUhs
and tJie included angle of one are, cqii/il, rcspuctinhj, io l/ro
adjacent aides and the inchidvd a>i;/l(.' •■fihi' uthn-.
Given ihe CD ABV}^i\m\ A'li'Ciy uwyVwh A ti = A' IV,
AD = A'I)', and lA =IA'.
To prove LTJ ABCD = CI7 A'B'C'D'.
Proof. Apply the £17 A'B'G'D' to the CJ ABCT) so that
A'D' shall coiueide with its equal AD.
Then A'B' will take the direction of AB (for IA'=IA);
•and point B' will fall on B (for a'B'=ab).
Then B'C and BC will both be 1 1 .-ID and will both
pass through the point B.
:. B'C will take the direction nf BC, Geora. Ax. .■!.
(thro)igh a gircn j>oiiit one siraighi line, ami oiilij one, can he {Irairii \\
aiiolktr gh'cn stmiglit liii':}.
In like manner, !>'€' must take the direction of />('.
.-. C must fall on C, Avt. C4.
(tuto siraighi lines can iittersci-t in hiil one Jioint].
.: Cn ABCI) = CD A'B'diy, Art. 47.
(geometric figures it-Mch eoiiicitie are eqval).
Q. E. D.
163. Cor. Two rectangles which have equal bases and
equal altitudes are equal.
Ez. Construct exactly an angle of 30°,
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164. A polygon is a portion of a [<lane boutitled by
Straight Hues, as ABCDE.
The sides of a polygon are its bounding lines; tlio
perimeter of a polygon is the sum of its sides; the angles
of a polygon are the angles formed by its sides; the
vertices of a polygon are the vertices of its angles.
A diagonal of a i)o!ygoti is a straight line joining two
vertices whkdi are not adjacent, aw IIT> in ¥\^,. 1.
165. An equilateral polygon is a jiolygon all of wliOKe
sides are equal,
166. An equiangular polygon is a polygon all of whose
angles are equal.
What four-sided polygon is equilateral but not eqiii-
angular ? Also, what four-sided polygon is both eituilateral
and equiaugular ",
167. A convex polygon is a polygon in which no side,
if produced, will enter the polygon, as ABODE {Fig. 1).
Each angle of a convex polygon is less than two right
angles and is called a salient angle.
168. A concave polygon id a polygon in which two or
move sides, if prudiK^ed, will enter the pob^gou, as FGUIJK
(Kg. 2).
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/4 HOOK T. PLAXE r.EOilEI-llV
Some angle of a ooneave polygon must be greater than
two right angles, as angle GlII of Fig. '2. Such an angle
is termed a re-entrant angle.
If the kind of polygon is not speui&ed in this respect, a
convex polygon is meant.
169. Two mutually equiangular polygons are polygons
wliorii- eoi't-espoiiding angles are equal, as Figs. 3 and 4.
170. Two mutually equilateral polygons are polygons
■whose corresponding aides ai'e equal, as Figs. 5 and 6.
From Figa. 3 and 4 it is seen that two polygons may be
mutually equiangular ■without being imituaily equilateral.
What similar truth may be inferred from Figs. 5 and 6f
171. Names of particular polygons. Some polygons are
used 50 frequently that special names have been given to
them. A polygon of three sides is called a triangle ; one of
four sides, a quadrilateral ; one of five sides, a pentagon ; of
six sides, a hexagon; of seven sides, a heptagon; of eight
fiides, an octagon ; of ten sides, a decagon ; of twelve aides,
;i dodecagon ; of fifteen sides, a pentedecagon ; of » sides, an
n-gon.
Ex, 1, Let the pupil illustrate Arts. 169 and 170 by drawing two
pentagons that are mutually equilateral without being mutually equi-
aDgalar, and another pair ot which the reverse is true.
Ex. 2. Can two triangles he mutually equilateral without being
mutually oquianguiar ? What polygons can f
Ex. 3. How does the iiumbet of vertices in a polygon uompare
witlj the number of aides !
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polygons '■>
Proposition XSXVIII. Theobrm
172, The sum of the angles of any polygon is equal to
iipo right angles taken as many times, less two. as the poly-
gon has sides.
Given a polj-gon of n sides {the above polygons of 5,
6, 7 sides being used merely as particulai- illustrations, to
aid in carryicg forward the proof) .
To prove the sura of its A = (ji— 2) 2 rt. A.
Proof. By drawing diagonals from one of its vertices
the polygon is divided into {n — 2) triangles.
Then the sum of the A of each triangle = 2 rt. /^ . Art,i34,
(lite mm of the A of a A w eqmt to 3 rl. A ).
Heiieethe sum of the Aot the {h— 2)&= (n — 2) 2 rt. .£ .
Ax. 4.
But the sum of the A of the polygon is equal to the
sum of the A of the {ii — 2)^. X-a. S.
Ileaee the sum of the A of the polygon = {« — 2) 2 rt. ^ .
173. Cor. 1. Thi> sum of the angles of a polygon equals
in— 4) rt. A.
eqiiiangidar polygon of n
rt. A.
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7() ]iOOK I. PLANK GEOMl/rnV
Pkoi'OSITIOX XXXIX. TuF.oiiKM
175. The sum of llif cxlerim- idi'jU-s of h pohjijoii formeO
hif pro'liMinil it.< siik.'i in ^sun-r.-'.'i'ii/ ill cif i:.flri:iiii!ij fijiitii.-
four right a>Hjk,
Given a polygon of n sides hiivlng iis buIcs pi-oiliiced
in suceession.
To prove the sum of the exterior A =4 rt, A .
Proof. If the iiiterioi" A of the polygon he denotod by
A, B, C and the corresponding exterior A by «, h, <\
lA-^ lfi^2 rt. A, (VThj!)
l!i + l}) = 2 rt. A , C^'hy ?)
etc.
Adding, int. A +ext. A = w time.'; 2 rt. :i =2jM-t. ^ . Ax. 2.
But int.i = ((/— 2)2rt.:i=2» vi.A — \vi.A. Art. I7:i.
.-. Ext. i=4 rt, A.
Q. E. D.
El. 1. What does the aum of the interior angles of a lieaagon
equal ? of a heptagon t of a decagon '.
Ex. 2. Each angie of an equiangular pentagon eontaina how many
degrees ? of au equiangular hexagon ! octagon ? decagon !
Ex. 3. Would a quadrilateral constructed ot rods hinged tit the
ends (i. e., at the verticea of the quadrilateral) be rigid ? Would e,
triangle so constructed be rigid J Would a pentagon (
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MISCELLANEOUS THE0EEM8
MISCELLAKEOUS THEOREMS
Proposition XL. Theorem
176. If three or more parallels intercept equal parts on
owe transversal, they intercept equal parts on every trans-
versal.
Given AP, P,Q, OR nnd T>T parallel lines intercepting
equal parts AB, BC and CD on the transversa! AB.
To prove that thev intercept equal parts PQ, QR and
liT on the transversal FT.
Proof. Throngh A, B and C let AF, BO and CR be
drawn parallel to PT and meeting the hues BQ, CR and DT
in the points F, 6 and H respectively.
Then
{twos.
In the
/ CDI{,
Also
Aut
But
{parallel ii
the lines AF, BO and CE are li , Art. 122.
aislU hues II a third straight line are || eai:h vther).
%. ABF, BCG and CDM, Z ABF = Z BCG =
{being ci-i. int. A ti/|| Unm).
IBAF^ ICBQ= IDCH, [Mmcreason).
AB=BC'=CD. Hyp.
.-. AABF=A BCG^ACBH. Art. 97.
.-. AF=BG = GR. (Whyn
AF^I'Q. liG=QR, CU = RT, Art. 157.
ifs 0("i;ii-f/(n(i;«[ bc(ii'p«i. pnraW lines arc equal).
.-, PQ^QR^RT. Ax. I.
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PnOPOSlTIOX XIJ. TllEdRElt
177. The line xvUdi join^ flw mhlpnhiU of livo shUs of
a triangle is parallel to the third nidu, and is equal io one-
half the third side.
Given J) the midpoint of AB, and J-J the midpoint <
AC in the triangle ABC.
To prove BE \\ BG and =iBC.
Proof. Through B let BL bo drawn 11 AC, and meetir
DE produced at L.
Then, in the A ADE and BDL,
AD^BD,
(Why!)
lABE^ZBBL.
(Wliyr)
ZDAE^ZDBL.
(Why?)
.: AABE^ABOL.
(Why?)
.: I'K^JIL, or DE^h BE, and A l-]---BL.
(Why?)
Bnt, A E ^EG {llyi^.) :. FA'.^-HL.
Ax. 1.
Al«o EC 11 BL.
Constr.
:. BLEC is ;i CJ ,
Art. 160.
' Ucu Hides of a quiulrUuleral arc equal ami ptirtdlH the figio
■eisfl^Z?).
:. BE 11 BC.
Art. 147.
Also LE^BC, {opp. sides ofaCy are =).
Art. 155.
.■.iLE,OTBE = iBG.
As. 5.
q. E. I).
178. Cor. The line which hiseds one side of a triangle
and is parallel to another side bisects the third side.
Thus, given AD = DB and DE 11 BG, then will AE=E<J.
For, suppose a line FG drawn through A \\ BG,
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MISCELLA.MEOUS THEOIUIMS
79
Then the three parallels FG, DE, JiC will intercept
equal pai'ts on AB. Hyp.
,". they intercept equal parts on AC. Art. 1T6.
.-. AE^EG.
Proposition XLIT. Tiieckem
179. The line wkieh joins the midpoints of the legs of
a trapezoid is parallel to the bases ami equal to one-half
their smn.
Given the trapeKoiil ABCB, E the mklpoint, of the leg
AB, and F t\\e midpoint of the leg' CB.
To prove that a line joining E .iiul /'Is 1| AT) and BC
and = i UD+ BC).
Proof. Draw the diagonal BI) and take (r its mid-
point. Draw EG and GF.
Then, in" the A AfW, BG\\ABa\i(\ = A AT). Art. J77.
{tU line which Joino f!ic miilpnfiits nf dm ^iili-^ iif a i la [l Ihe (liwd stile
and = oiic-half tin: third eiile).
Also, iutlie Ai'i^C, GFllliCand^i BC, {sawc r.r^mn).
.: OF ami AD hoihwnC; .-.CFwAI), Art. 123.
{two striiii/Jii lines \\ a t/iirU stnUnht are \l each oi/iei-):
.-. EG and GF are both 1| AD.
• '. EG and GF form one and the some straight line EF,
( Oirmigk a giccn point one line, and onhj one, caa be drawi
gixcnImcS.
:. EF\\ABm(kBC.
Also EG = I AB, and GF=l BC.
Adding, EG + GF, or EF=^ {AB^BC).
180. Cor. A line draicn hi.'fecfh:
and parallel lo fhe base hisccln the oh
OilclPf, of,
r hij also.
mioliier
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]iOOt; I. rj-ANE GEOJIKIUY
Propositios XLIII. TuEOiiEir
Tlu- Inscrtors of thr am/kn 'fa trlimt/ir
Given Uie A ABC with tlic lines .1/', BQ, CJi (Fig- 1)
bisecting the/. BAC, ABC, ACB, respectively.
To prove tlmt AP, BQ, CR intersect in a common point.
Proof. Let AP, the bisector of / BAC, aiul CR, tlie
bisector of / BCA, intersect at the point 0.
Then 0, being in bisector /IP, is equidistant from AB
and AC, Art. 117.
(every point in the hmctor of nnLU equidistant from Hie tides of IheZ).
Also 0, being in bisector CR, is equidistimt from .If" nncT
Ii( ', {mini! iT'fXoii).
llenuc O h cqiiiOistaiit I'roni tlie riidus ,-l/>' anJ li< '. Ax. 1,
. -. (I is in the lilsector of /. A BC, Art. i it.
{every point e^idiliMiintfrom the sides of'luZ lies in the hiMVtor of Qia Z.) .
Hence BQ, or BQ prodtiueil, passes through 0.
Hence the biseetors, AP, BQ, CR, of the three A of the
A ABC intersect at 0. Q, E. D.
182. OoR. The point in which the three bisectors of the
angles of a triangle intersect is equidistant from the three sides
of ike triangle.
1 83. Dbf, Concurrent lines are lines which pass through
the same jioint. _„_____
Ex. Find other Z of above flgurs if Z /M f = 73= aud Z BCA =44'.
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MlSCELLANEOrS THEOREMS
Proposition XLIV. T
184. The perpe-ndicular bisectors of the sides of a trian-
gle intersect at a common point {called the circumcenter).
Given tlie A ABi- with DP, EQ, Fli. (Fig, 1i, Uio j_
biseclufa of tliu sides AB, H<.\ C.\, reKpectivoIy.
To prove tiiat T)P, EQ, FR intersect at a comimin point.
Proof. Let DP and EQ intersect at the point 0.
Then 0, being in _L OP, is eqnidistiiut from tlie jjoiots A
and B, Art, 113.
{ever-ff poiiU iii, We perjieiulkiilttr bisector of a line in equaily dieivnifrmn
ilte exbremiiiee of ike Urie) .
Also 0, being in J_ EQ, is eqnidistunt frorii the points
B and C,
(mme reawu).
Hence O is cquidist[uit fram .1 imd ( '. Ax. 1.
.-. is in the ± bisector of AC, Ait. 115.
{the, \_ hiaector of u line U Hie locus of alt points equidisUfiil from the
exireifiitiei! of the line).
Henee FR, or FR prodnoed, passes throngli ().
Hence the perpendicuhir bisectors, }iP, EQ. FR. <i\' ibc
lliree sides of the A ABC meet in tlio point 0. a. e. a.
185. CoJ!. '!'}i<- pi)i))l ill which lin; pi'r}n'iidivuhir hiKi'vlKn^
of Ih,; .u(l,s of a iriuiujl, mcvl is equidisloiit from thi- verliim
of the triaiujU:
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[■LANK GEOMF-TRY
rilOroSITION' XLV. THEORE^il
186. Tlif perpi'mVculars from the ret-firca of a triangle
to the oppogi/e sldca iiuet ut a point {called Ihe ortho-center).
'A
N
M
/^
w
Given AD, BF, am\ CB the perpendiculars from the ver-
tices A, B, and C of the A ABC to the opposite sides.
To prove Ihiit AD, BF, and CE iotorsect in ;i comiiiou
point.
Proof. Through A, B, and C let the lines PR, VQ and
QR be drawn || BC, AC, and AB, respective! y, and forming
the A FQR.
Then AD 1- VB, Art. 123.
(/,»■ AD 1 BC. njjrf r, U:ir J. n:ii; of lli-o [[ !i»rs U 1 the Oilier ulm) .
Also A ]' BO imA A BCli are Z17 . Constr.
.-. AT = BC, and AB = EC. Art. 155.
(((((! opponiic si(}es of a CD are = ).
.-. AP = AR. As. 1.
lience, in the A FQR, AD is the perpeudicukr bisector
of side PR.
In like manner it may be shown that BF is the perpen-
dicular bisector of FQ, and that GE is the perpendicular
bisector of QR.
Hence AD, BF, and CB are the perpendicular bisectors
of the sides of the A PQB.
.: AD, BF, and CE meet in a conimon point, Art. 184,
ithi }>ci-pendieula,r biseeiois of the sides of a A arc concurrent).
Q- E. B.
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miscellaneous theokems 83
Proposition XLVI. Theorem
187. The medians of a triangle intersect {or are con'
current) in a point {called i?ie centroid) which cuts off two-
thirds of each median frotn its vertex.
Given AD, BF and CB the three medians of A ABC.
To prove that AD, BF and CB intersect at a point which
cuts off two-thirds of each median from its vertex.
Proof. Let the medians AD and CE intersect at 0.
Take R the midpoint of AO, and S the midpoint of OC,
and draw RE, ED, DS, and SR.
Then ED 11 AC and = I AC. Art. 17T.
Also, in the A AOG,
lis II AG and - i AC. Art, 177.
ED II RS and = ie^'. Art. 122 aad Ax. 1.
.-. REDS is a £3' .
ES and RD bisect each other.
AR^BO, and CS=^SO.
.: AR = RO=OD, and CS^^SO^OE. Ak. i.
Hence t'£erosses.'IJ.'at a point 0, such that A0=|.4i).
In like manner it may be shown that BF crosses AD at
the point 0.
Uuuite the medians A D, BF, iuid OF intersect at point 0,
which cut-s otf two -thirds of each median from its vertex.
Q. E. D.
Hence
Hence
■Bnt
(Why!)
(Why!)
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84
BOOK I. TLAKE Cr.OHETEY
188. Properties of rectilinear figures to be proved by the
pupil. Proof of equality of triangles. Otlier properties of
reetUioear flgares will now be given vvliioh are to be demon-
strated by tlie pupil. These theorems wili be arranged in
groups, according to the method of proof to be used,
followed by a group of general ur mixed exercises.
Le.t the pupil form a list of the, conditions that make
two triangles equal.
(See Arts. 96, 97, 1)8, 101, 102, 141, 142.)
£XERCIS£S. CROUP 4
EQUALITY OV 'rilIA.\'Gl.ES
Ex, 1. Riven AHC any triangle, BO Iha hi- £
seotor o( I ABC, and JD 1 BO; proved J B0==
A BOD.
[SUG. In the A AffO and nno wbal Hues
are equal! What A are equal t etc.]
Ex. 2. If, at any point in the biseolov of
an^ie, a X be ereeted and produced to meet t
Bides of the angle, how many triangles bvo forint
Are these triangles equal ? Prove this.
Ex. 3. If, through the midpoint of a f
straight line, another line be drawn, and prnduc
to meet the perpendiculars erected at the o(
of the given line, the triangles so formed i
P^
sect each other and their
of equal triangles are
Ex. 4. H two straight 1.
ties be joined, how many
Prove this.
Ei. 5. It equal segments from the base be
of an isosceles triangle, and lines be drawn from
segments to the opposite vertices, prove that t
angles are formed.
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ESEECISES. EQUALITY OP TRIANGLES 85
Ex, 6. If, upon the sides of an angle, equal segmnnts be laid off
from the vortex, and lines be drawn fram the ends of these Begments
to any point in the bisector of the angle, prove that the trianglea
formed are equal.
Ex. 7. If two sides of a triangle be produced, each its own
length, through the vertex in whioh they meet, and the extremitiea
of the produced parts be joined, prove that a new triangle is formed
whieli equals the original triangle.
Ex. 8. Given AB^DC, and BC^DA; yro-
ABAC=A1>AC. W
triangles is tbare in the figure
, and BC^ DA : prove /\g-^
Ex. 9. Two right triangles are equal if ilieir corresponding legs
Ex. 10. The ttltitudsH from the extremities of the base of an
isosceles triangle upon the legs of tlie triangle divide the tiguve into
how many pairs of equal triangles ! Prove this.
Ex. 11. In a given quadrilateral two adjacent aides are equal and a
diagonal biseots the aogln between tlie^e .■sides. Prove tliat the diagounl
bisects the qnadrilaternl.
Ex. 12. If, from the ends of the shorter base of an is os eel es trape-
zoid, lines he drawn parallel to the legs and produced to meet the
other base, prove that a pair of equal triangles is formed.
189. Prooi of the equality of lines. Thei-c are several
methods of proving tiint two lines (segments) are equa!.
One of the principal tnetJiods of proving that two lines are
equal is J>y provitifj that two triangles, in tvMch the given
lines form homoloijoit» /uirls, are equal.
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EXERCISES. CEtOUP a
EQUATiiTY OK j.ixr:.-;
Ex. 1. Given ABC niiy t.-ismsle, JIO the h)-
BPPtorof lABC, and -J/) ± HO; prove AO = On.
Ex, 2. K, at any point in tbe bisector of so
angle, a porpendicular be ereeted to the bisector
and produced to meet the sides of tho angle,
the perpendicular 13 divided into two equal parts at
Ex. 3. If two aides of a triangle be produced,
each its own length, tbronyh the common vertex,
the line joining tbe extremities of the produced
parts equals tbe third side of tbe triangle (i e.,
!»£ = BA).
Ex. 4. It eqnal segments from the base bi- liijil
otT on llie legs of an iaoectles triangle, lined dr;i«n
these segments to the opposite vertices are isqual.
Ex. 5. Given AR\\ QB, and AF = FJ1:
prove BF = PQ.
Ex. 6. The bisector of the vertical an-
gleof an isosceles triangle bisects the base. '^ ■"
Ex. 7. Tiie aititiides of an isosceles triangle upon the
eqnr
rectangle are equal,
isosceles triangle to tho legs
nt a triangle are equal, the 1
Ex. 8. The diagonals of
Ex. 9. The medians of i
Ex. 10. If two altitude-
ioaeeles,
Ejt. 11. Tbe perpendiculars to a diagonal
f a parallelogram from a pair of opposite
ertiees are equal.
Ex. 12. If the equal sides of an isosceles triangle be prod
hrough the vertex so that the produced parts are equal, the lines
□g the extremities of the produced parts U> the extremities o
lase are equal.
Ex. 13. If tbe base of au isosceles triangle be trisected,
IrawL from tho vertex to tbe points uf ti'iaectiun are equal.
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EXEECISES. EQUALITY OF ANGLES
87
Lines may also be proved equal by showing that they are:
opposite equal angles in a triangle;
or opposite sides of a parallelogram;
at parallel lines comprehended hetweeit parallel lines, etc.
(See Arts. 100, 155, 157, ete.).
Ei. 14. If the exterior angles at the base
of a triangle are equal, the trian-
Ex. 15. In the biseotora of
the equal angles of an isuseftles
triangle, the segments nest to
the base are equal {AO= OP).
Ex 16. In A ABC, given
AJ) = AC. DE II BC; prove
Ah = AE.
Ex. 17. Given AB = DC,
a.)iABC = AD; provf, AE= EC.
190. Proof of the equality of angles may be obtained
in several different ways.
One of the principal methods of proving that two angles
are equal is by proving that two triangles, in tvMch the gieen
angles form Jyyimlogous parts, are equal.
EXERCIGES. CROUP' &
EQUALITY OF ANGLES
i1. 1- Given J^r: any A, BO the bisector
of the lABC, amlJDX/iO; prove lliAO
= ^ BDO.
Es. 2. If, at any point in the bisector of an
angle, a perpendieular bu erected and product;J
to me^t the sidea of the anjtle, the ptr-
pendieular makes equal angles with the sidett
of the angle.
Ex.3, Given AH-liV, M^ AD - BV;
pruvb IM~/ n.
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88 IJOOK I. I'LANE GEOJILTJIV
Ex. 4. If equal segments from the base be laid olT on Iho li%'3 of
an isoswlea ti'iaiigle, the lines <lrawn from the extreniilips of ibe
segments to the opposite vevtii;es miike equal anglus vvilh llie base.
Ex. 6. The altitudes upo
equal angles with tbts base.
Ex. 7, Theai.tgoiia
Aiij^los may iilso he proved cqivA
aft- opjMsite equal sides in an i^oKcrli-i^ li-ianyh;
or uir veriical angles;
or ore complements [or supfiUmenlx) of equal ani/les;
or hij the use. of the propeHies of parallel lines;
or that their sides are parallel, or perpendicular.
(Sco Arts.
78, 99, 124, 1120, 130, 132.;
I isosBfiles ti-ia,ngle the txteiior auf^li
■ing the base are equal.
E*. 9. Given AC = CB, an.t DE
Jill; prove Z.CDE = ZCJSn.
',x. 11. Conversely given
= Z B, and CE \\ AB ;
■e that CE bisects I DCA
Ex. 12. Given BD tbe bisector of thi
angle ABC, and PR \\ CB; prove PJIB
isoseelea A. Let the pupil state ibis theuri
in general language.
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EXEKCISES. PAEALLEL LINEH
hig ihe a'ier-
Ex. 13. Given AB = AC,
aii3BD=CE; prove ZBCfl=
Z CBE. How many pairs of
equal A in the figure f of
equal lines? of equal A1
Ex. 14. A line drawn
through the vertex of an angle,
perpendicular to the bisector
of the angle, makes equal
angles wilh the sides of the given angle.
Ex. 15. If a Btraigbt line which biaecis
one of two vertical angles be pi'oduued, it
biseots the other vei'tioal angle also.
191. Proof that two lines are parallel i
bj' showing that:
itie Ihu'fi lire cul ha a l7-(ntsi'ernfil, »
nate interior angles equal;
or making the extenor interior angles equal;
or making the interior angles on the same sitle of the
transversal supple^iieniary ;
03- that the lines are opposite sides of a parallelogram;
or tJuit one of the lines joins the midpoints of tico sides
of a triangle, and the other line is the third side of
the triangle.
(See Arts. 125, 127, 129, 147, 177, ett.}
EXERCISES. CROUP 7
PARALLEL LINES
Ex. 1. If two sideg of a triangle be produci'ii, each
vertex, the line joining their ex-
tremities is parallel to the Ihird
side of the triangle.
Ex, 2. The bisectors of two
alternate interior augles of pui-
allel lines are parallel.
A = ZS, and IDCE = lECA ;
-Ex. 3. Given
rove CE \\ BA.
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\n\ HOOK 1. PLANK iu:o.Mi/n;v
Ek. 4. The hi life tors o£ the opiiositci aiiflea oC a ]Kir;ilkloi;ram are
Ex. 5. l.im.a perpcu.li^iilar lo parallel Imp-. Krs pamllel (or
B paiiillfl if Iwo points on oua iiue
192. Tlie proof of a numerical property of the recti-
linear figures (of Book I) usually depends on one of the
following;
The sum of the angles abmit a given point on the sami-
side of a straighl line passing through the point is 180 ;
or the smn of the angles about a point is 3G0°;
<ii' the sum of the angles of a triangle is 180° ;
or the .iiiiu of the interior angles of parallel lines on the
same side of a transversal is 130°;
or file sum of the interior angles of a polygon of n side^:
is (ft--2) 180°;
:;i- the Slim of the erlertor angles of a polygon is 360°.
(See Arts. 76, 77, 128, 134, 172, 175.)
EXERCISES. CROUP 8
^;L■.■^[EKK'AI. rROl'KKTlKS
Ex, 1. It an e-ttfirior angle o£ a triaiitcle is "123° and an opp^
inferior angle is SK", find the other two angles of the triangle.
Ea. 2. Find the angle formed by the bis'eotoi-B of the two s
bugles ot a right triangle,
Ex.3. If two angles of a triangle are 50" and 60", find the s
formed by their biaaotora. Find the same if the two angles toi
Ex.4. It -he vertm aiigle of an irascejes triangle is 4(1" a
;pei'pendieiilar I. drawn from an extremity o£ the base to the opp
■liile, fliiU Ibe ai.^^lea of the figure.
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EXERCISES. NUMEKICAL PROPERTIES
Ex. S.
If til.- v.rfti'X aiife-le of an isosceles ti-iangl.)
^40"
find tl
^le me
tidfd between the altitudes drawn from tlie
tiie oppoiiite sitles.
Bxtve
mitiea o
Ex.6.
How many degreaa in eac:h angle of an equiu
ij:;ula
doiieea
n? of
n equiangular li-gon?
Ex.7
How many diagonals are there in a pcntaf-o
decagonT inanx-gon!
uT i
a hexn
The methods oi proving that a given angle is a right
angle (or that a given line is perpendicular to another given
line), or that one angle is the supplement of another, -aiv
closely related to tlie ubore tHethuiU of ubtiiinhuj llw iiiiiHi-ri'
(■(tl v(il„.''8 of (jivi'H iiiiglef<.
Ex. 8. .\i.y iniir of adja.'.^iit aiigleH of ;. parallelogram is mipplu-
Ex.9. If oneaugleof aparalie
a rectangle.
Ex. 10. The bisectors of two (,
adjacent angles form a riplit angle
ograin is a rigli
upplementary
(ave verpOii-
iterior angles o
angle the fignre is
Ex. 11. The biseetors of two i
the same side of a
Z.^ form a right angle.
Ex. 12. If one of the It'gs
(JB) of an isosceles triangle
he produced its own length {ISD) and its ex
tremity (D) bo joined to the other end of the base
(C), the line last drawn {DC) is perpendicnlar to
the base.
193. Algebraic method of proving theorems. The proof
ci certain properties of a geometric ligiu'ti is ut'teii fin-ilitiitfil
ty the use of an tilgehraic si/iiihoJ for un Kiil.iioini aiKjh "i-
tm itjik'iiown line of the. fiifKir. (i>id tin' iifit of uit i<jutilion or
ether ulyebrak metkud t/J solution.
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;)li I'.OOK I. I'LANE (iEUMlITKi'
EXERCISES. CftOUP 3
Al,i;EHtiAIO Mirnillll
Ex. 1. Find the number o[ degrees in :iii an^le wlii'.'h eqiiiils twice
its complement!
[Srti, Let ^ = 1116 complement, etc.]
Ex; 2. Fiud the number of degrees in nn nugle which equals its
supplement? in one whk'h equals oue-third its supplemeut!
Ex. 3. The angular space Kbout a point is divided into four angles
TObieli are in the ratio 1, 2, 3, 4. Piud the number of degrees iu eaeU
angle.
[Sl-o. j:}- 2.i+:!i--t-4J^ = 3tiO°, etc.].
Ex, 4. The ougles of a triangle are in the ratio 1 , 2, a ; lind the
angles,
Ex. 5- Two angles are supplementary and the greater exceeds the
less b>- ^O" ; find tlie angles.
Ex, 6, Find all llie .angles of a paraUelogram it one of tiiem is
double iLiiother angle.
Ex. 7. One of the base angles of a triangle is double the other, and
the exterior angle at the vortex is 105°. Find the angles of the triangle.
Ex. 8, How many sides has a polygon the aum of whose angles is
fourteen right angles?
[Sl-g. 2(h-2) = U; fimln.]
Ex. 9. How many sides has a polygon the sura of whose angles is
ten riglit angles! twenty right angles! 720°i
Ex. 10. How many sides has an equiangular polygon one of whose
angles is seven -fourths of a right angle f
Ex. 11. How many aides has a polygon the sum of whose interior
angles equals the sum of the esterior angles !
Ex. 12. How many sides has a polygon the sum of whose interior
angles equals three times the aum of the exterior angles?
Ex, 13. If the base of any triangle be produced in both directions,
the sum of the exterior angles thus formed, diminished
by the vertex angle, is equal to twi right angles.
[Sua. 180°— a-i-lSO'^— ft-tlSO"— a— (i)=,6te.]
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EXERCISES. AUXILIARY LINES
Ea. 14. TliO bisectors of the base angles of a
iaoscelea triangle include an angle -which is equii!
the exterior angle at the base.
[Sm. To prove « = '!>, denote one of the baso A.
2x, etc.]
Ex.15 In an i-osccles triangle the altitude upon
one of the legs makes an angle with the base which
eqaala one-half the lertcx angle.
[Suu To proie «^ Wi, show that n = 90°— r, i = lSO°
Ex. 16. If the opposite angles of a qiiadri
lateral are eqaal, the figrure is a, parallelogram
194. Use of auxiliary lines. laequalities. The demon-
stration of a property of a geometrical figure is frequently
facilitated by drawing one or more auxiliary lines on tlie
figure. For examples of the nse of such lines, see Props.
Ill, V, IX, etc., of Book I.
Some of tiie principal auxiliary lines ii!>eil on reutiUucar
figures are:
a line connecting iwo given points;
a line through a given point parallel io a (liven Vin< ;
a line- through a given point perpendicular to a gicen line;
a line making a given angle with a given line;
a line produced its own length, etc.
EXERCISES. CROUP
AfXILlARY LINES
2i£. 1. In the quadriiaterul
AB^AD, and 110 = 0); prove
[SuG. Draw AC, etc.]
Ex, 2. Prove that the nn(;h
of an isosceles trupezoid art S'
IJI = / II.
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04 BOtHi I, I'L.VKK (IKUilETKV
Ek. 3. titiile and prove the converse of Ex. '1.
^ El. 4. GivRii J JS jl ClJ ; provR z ?i = Z ,<
^-■>,,^^ Ex. 5. Conversely, Kiveti Z ?^
'^ — "■'^ = Z ,1 + Z c ; prove JH |1 CD.
Ex. 6. Tlio median to the hypotenuae at a ri;;!:!
tiiaiiglti is oae-hal£ the hypotenuse.
Ex. 7. If one acute angle of a right triangle is double
the other, the hypotenuse is double the shorter leg.
[Suu. Draw the median to the hypotenuse, etc.]
Ex. 8. In an isoBeeles triangle,
the Slim of the perpendiculars
drnwii from any point in the base
I to the legs is equal to the altitude upon one of
U the legs.
la some cases it is useful to <iraiv firo <
Ex. 9. Show that the median of a trapezoid equals one-half the
siun of Ihe two bases by drawing a line through the miiipoirt of one
IfC of the trapezoid, parallel to the other leg and meeting one base
aud the other base produced.
Ex. 10. Lines joining the midpoints of the sides of a quiidrilatei-ni
taken in order form a parallelogram.
[Sui!. Draw the diagonals of the quadrilat-
eral and use Art. 177.J
Ex. 11, If the opposlti
sides of a hexagon are equa
and one pair of sides (.4B ami jj;
CD) are parallel, the opposite anglcB of the he.-tagoi
Ex. 12. 'Hteu Allelic, and An=CJi; prov
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EXERCISES. INEQUALITIES Vii
Let the student form a list of tlie principles proved in
Book I concerning unequal lines ottd unequal angles.
(See Arts. 92, D3. 95, etc.).
EXERCISES. CROUP 19
INEQUALITIES
Ex, 1. In the triangle ABC let i' be any pnint in the Bide BG;
prove All + BC > AP '.- PC.
Ex. 2. In the quadrilateral Alscn let F be any point in the side
.Wf; prove perimeter of APCD > perimeter o! AfD.
Ex 3. In the triangle AISC let 1) he any point in AB nnd F any
point in IW. Ftove AB + BC > An + I)l'-+ FC.
Ex. 4. The sura of the four sides of a (piadri lateral is g.eattr IIihu
the Bum of the (ilagonala.
Ei, B. If, from any point within a triangle, liiiea be diaivn to the
verfiees, the sum of the hues drawn is greater tbau one-half the sum
of the sides of the triangle.
[SUQ. Use Art. 92 three times, etc.]
Ek. 6. If, from any point, within a triangle, lines be drawn to the
vertices, the sum of the lines drawn is less than the Bum of the sides
of the triangle.
[Sue. Vse Art, 05.]
Ex. 7. The median to any side of a triangle
leas than half the sum o£ the other two sides.
[SiTG. Prodnee the median its own length.] j
Ex, 8. If B is the vertex of an isosei'les triangle vlBr, r.
produced to the point J), lliAti ia greater than IBIIA.
Ex. 9 lu the figure 51. 7],7t(.' > AB; prove llitV yr*
/.BVA. [Sifi, Use Art. 108 }
Ex. 10 111 the same figure, show that /■'(' < /.'' .
Ex. 11. Ill the qimilrilalet-al
liV the shortest : prove ,'- .(/((-srej
llian I BAD.
Ex. 12 Lines are drawn from .i, ff, ('. ?i, Coiirpoinls in a
line, to the point f ontside of the line Wlucli anj;lo3 on 1
atu lefiS than angle ACF^ Which angles are greater I
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1 '1 1 BOOK T. I'T.AXE GP:OMETI{Y
195. Indirect demonstrations. Loci. Tii Book I tliri'e
nnjthnds of indirect proof hiive been used.
1. The reduction to an absurdity {rednctio ad absiiv-
dum), that is, the proof that the negative of a given theorem
leads to mi ubsuvdity Csee Prop. 5).
2. The method of exclusion, that is, showing that any
other statement ihtn the. i/iven theorem cannot he true (see
Pi-ops. XV, SVII). This method is a special ease of the
preceding, the negative of a given theorem being divided
in it into ti^'o parts ■which are separately shown to be
impossible.
3. The method of coincidence, that is, proof that a given
line coincides idth annthrr line, irlii'-Ji fulfils cvrioin re-
quired cmiditions (see Props. XVII, XXIIJ, XXV, etc.).
EXERCISES. CROUr J3
INDIRECT, OE NEGATIVE DEJIONSTKATIONS
Prave the folloiyiiig by iiu uvlitneX method:
Ez. 1. Every point within an angle and not in the liiseetor of the
angle is unequally distant from the sides of the an^ic.
[Si,"Q. In the given angle take P any point not in llie bisector of
the angle. Then, if P is not unequally distant from AG and on, it
must be equally distant from them, etc.]
Ex. 2. I£ two straight lines are cut by a transversal, maldug the
al:ernate interior angles nnequal, tie lines are not parallel.
Ex. 3. The line joining the midpoints of two sides of !i triangln
is parallel to the thii'd side.
[Sua. Through one of the midpoints draw a line |1 to the Ihini
side, show that it bisects the seeond side and that the line joiulug the
midpoints eoinaides with it. J
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EXERCISES. LOCI
97
Ex. 4. IE, ffom a point P in a line AB, Jines PC and FD be drawn
on opposite sides of AB making the angle AFC equal to tiie angle BPD,
PC and PD are in the same Htraiglit line.
[SuG. From P draw PQ in the same straight line with PC and
show that PD eoineideB with it,]
Ex. 5. The hlaectors of two vertical angles are in the same
straight line.
Ex. 6. In the triangle ABC, Bis any point in the side AB, and Eis
any poi
Ex. :
a given
II tho side
Proi
i that BE and IJC cannot bisect each
Whnt
Vi-<
EXERCISES. CROUP 1-3
LOCI
he iiifus of iill points :d a pivf
e this.
Ex. 2. What is the locus of all points equidistant from two given
parallel lines f Prove this.
By use of known loci (see Arts. 112-118), prove the following:
Ex. 3. The diagonals of a rhombus are perpendicular to each
other. [Su«. See Art. 113.]
Ex. 4. The median of tin isosceles triangle is perpendicular to the
Ex. 5. The line that joins the vertices of two isosceles triangles on
the same biiRe is perpendicular to the base.
1 96. General method of obtaming a demonstration of a
theorem. Analysis.
A due to the sointion of some of the raoi-e difficult
theorems is often obtained by proeeeding thus:
Assume the proposed theorem as true; observe tchai other
felaiion among the parts of the figure must then be true;
proceed backit-ard thus, step by step, till the required theorem
1^ found i-o depcwl on some linoirn truth; then, starting with
this knoivn truth, reverse the ste/is tiilrn, (iiid thus build up
a direct proof of the nqnireil theorem .
This method is called solution by analysis.
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98 BOOK 1. PLAXE GEOMETRY
The following is a simple example of (lie
method :
Ex. Given AB and AC Hie logs of an ienacoloa
triangle and D any point on AD; prove DC greater
than DB.
Analysis. If DC > DB,
I Bis greater than I DCB. Art, 104.
Hence subatituting for ZB its equal, lACli, we
have Z^C/; is greater than ^DCB, Ax. 8.
But we know that lACB > IDCB. As, 7.
Hence, Direct Proof (or Synthesis)
ZACB is greater than ITlCn, Ai. 7.
.-. Z B is greater than IBCE. Ax.. 8.
:.DC > DB. Art. ICU,
0- E. D.
The first part (aualysis) of the above process is to be
purely mental work ou the part of the pupil, in investiga-
ting a given theorem; the second part (the direct proof, or
synthesis) is to be written out as the required solution.
In working the following exercises, this method will be
found to be necessary in the solution of only a few of the
more difficult theorems.
EXERCISES. QFiOUP 14
THEOREMS PROVED BY VAKIOUS METHODS
Ex. 1. If two opposite
diagonal connecting thei
tills diagonal.
;a of a quadrilateral are blaected by the
ices, the quadrilateral U bisected by
Ex. 2 Perpendipuiars dras
triangle to the median to the b
a the extremities uf the base of a
Ex. 3. If the perpendiculars from the extremities of the base of a
triangle to the other two sides are equal. (1) these perpendiculars make
«qual angles with the base, (a) the triangle is isosceles.
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MISCELLANEODS EXEItCIBEH
Ex.4. I( tlie lines Jfi and CD iulorse
rsect.tbon Ali+Cn > AC'+DB.
iviAes trLacgle is 44^, and one ot
praiiuced to me el the opposite
Ek, 5. The verlPK aii)|l« cf an u
the base anplea is bisected by a hn
side. Find ad the angles of the figu
Kx,6. In the figure of Prop, V prove that ZJPC is greater than
I ABC. Also prove the same m another wiiy by means of an auxiliary
line drawn througli fi and F,
Ex. 7. Ptrpeudioiilnrs drown from the mi.lpoinl of the ba.^ie of a:a
iaosoeles triangle to the leRs are etinal.
Ex. 8 State and prove the
Ex. 9. In an Isosceles triat
Ex 10. In a re ent
angle at the re entrant
t quadrilateral the exterior
tes equals the sum of the
dueedtoD; prove BD > JD > JB,
Ex, 12. If irOLn a point in the liisector ul au obliqiic aiiglu
formed is a rborabus.
Ex 14. It the median nf a triang
the triangle is isosceles,
Ex. 15. Given JB = Ja
lBAC = ilB,
and DFX ISC,
prove A EFA equilaleral.
, Unalvsis. 1{ a fJA-i>equilatei
= dO^
Her
"sing iLbAC= ilC.\
% din
U, lfEA = W°; .■.Z»Kfi = GO°;
proof, show that IC ~ 30° by
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1 00 BOOK I . I'iANE L"; KOMKTliY
Ex. 16. If Ihe ilingooals o£ a quadrjlatoral bisect each other at
right angles, what kiad of a figure is tlie quaJrilateval f Prove this.
Ex. 17. From the point in which the altitudes drawn to the legs ot
an isosceles triauf;le intersect, a line ia drawn to the vertex, Prove
tiat this lino bisects the attgle at the vertes.
Ex. 18. If from a point within an acuta angle perpendiculars ai'e
drawn to the sides o£ tlie angle, the angle formoii by thesa perpendictt-
tara is the supplement of the given angle.
Ex. 19. Linos joining tlio midpoints of tli
aides of a triangle divide the triangle into Ecu.
eqaa! triangles.
Ex. 20. If, in the parallelogram ABCD,
BP = VQ, then AQCF is a parallelogram.
[A>-.\LYsiS. If J^CPisa^:?, JP=find
II QC. .'. hegin the direct proof by sliow
ingthat JP=andi3 || QC]
Ex. 21. If the diagonals of a parallelogriim are equal, what kind
of a figure is the parallelogram! Prove this.
Ex. 22. If the cngle A of the triangle AUG is 50° and the oxterior
angle BCD is 1L'0°, which is the largest side in the triangle !
Ex. 23. Two triangles are eqnal if two sides and the median to
one of these sides in one triangle are equal, respeistively, to two
homologous sides and a median iii the other.
Ex. 24. Two isosceles triangles are eqnal if the base and an angle
of one are equal to the base and the homologous angle of the otiier.
Ex. 25. Two equilateral triangles are equal if an altitude of one
IB equal to an altitude of the other.
Ex. 26. If two medians of a triangle are eqnal, the triangle ia
isosceles.
[SuG. On Fig, to Prop. XL VI, taking AD= EC, prove A AOC isos-
celes, AAEC=l\dDC, etc, How could this theorem be investigated
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Miscellaneous exekcises
Ex. 27. Prove the sum of the aiif-lea of a 1
right angles by drawing a liue through the ve
parallel to the baae.
:ri:mglft equal to tw
rtes of the ti'iangl
Ex. 28. The homologous mediaus o£ two
equal.
equal triangUa ar
Ex. 29. The bifleetora of nn angle of a tri-
angle and of the twn erterloi- aiigios at the other
A
Ex. 30. If JKC is an equilateral triangle nTul
AP= BQ= CE, then I'QIt is an equilateral
triangle.
Ex.31. If the two baae ancles of ft trianglo
l)e bisected, and through the point of ii;tei'See-
tion of the two bisectors a line be drawn parallel
to the base, the part of this line intercepted
between, the two sides equals the sum of the
aegments of the sides included between the par-
allel and the base (i, e., prove PQ=AP+ QC).
Ex. 32. Two quadrilaterals are equal, if three sides and the two
included angles of one are equal to three aidea and the two included
anglea of the other, respective!}'.
Ex. 33. If the diagonals of a quadrilateral bisect each other, the
figure is a parallelogram.
Ex. 34. Lines joining the midpoints of tho sides of a Vtctangle
in order form a rhombus,
Ex. 35. Lines joiuing the midpoints of the sides of a rhombus
l^orm a rectangle.
Ex. 36, The bisectors of th? angles of a paraliclogram form a
reetangle.
Ex. 37. The bisectors of the angles of a rectangle funn a square.
Ex. 88. If lines be drawn through the v
parallel to tho diagonals, a parallelogram ii
M large as the original quadrilateral.
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1>I,ANE GHOMK'I'RV
grata to thtt luuipo'iit'i of a, pair of opposite sides trisect;
of the pavallelogram .
Ex. 40. Oq the liiaKOnal AC ot a paralli4osram AltCD ei)
AP and CQ, are marked oil, Prove Bl'HQ a jwi'allelogl'a
many pairs of equal friaiiglps does the ligiive i-oiitaiii i
Ex. 41. Tlie oppo!
Ex, 42. In an isosed™ trapezoid, tiie diagonals are equal.
Ex. 43. If the upper base of an isosceles trapezoid equals the eum
of the legs, and lines be dran-n from the midpoint of the upper base to
the exiremities of the lower bas^e, how mauj' iKoseeles triangles are
formed ! Prove this.
Ex. 44. The lines joining the midpoints of the sides of an isos-
celes trapezoid, taken in order, form a rhombus ov a squary.
Ex. 45. The bisectors of the angles of a trapezoid form a quadri-
lateral whose opposite angles are supplementaif.
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Book U
THE CIRCLE
197. A circle is a portion of a plane bounded by a
curved line, all points of wliiisli are equally distant from a
point within <mllod the center.
The circumference of a (linUe is the curved line bounding
the circle. The term itin^le may also be used for the
bounding line, if no ambiguity results.
A eii'ple is named by nomiiii; its center, an the eircile O; or by
naming two or more points on its cireumfereuce, ag tlio uirtlo AVD.
198. A radius o£ a circle is a
straight line drawn from the center
to any point on the circumference,
as AO. A diameter of a circle is a
straight line drawn through the cen-
ter and terminated by the circumfe-
rence, as BC.
199. An arc is any portion of
a circumference, as AG. A semi-
circumference is an arc equal to
one-half the eireumfei-ence, as
BAG. A quadrant is an arc equal
to one-fourth of a eireumferenee,
B.SBD.
200. A chord is a straight line
Joining the extremities of an arc,
as EF.
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104 BOOK Tr. PLASE (lEOMETKV
Every chord subtends two arcs. A miuor arc is the
smaller of two ares subtended by a chord. A major arc is
the larger of two arcs subtended by a chord. Thus, for the
chord EF the minor arc is EPF, and the majnr arc is ETF.
Conjugate arcs is a general term for a pair of minor and
major aros.
If the are subtended by a given chord is mentioned,
unless it is otherwise specified, the minor arc is meant.
201. A tangent to a circle is a straight line which, if
produced, has but one point in eonunon with the circle, aa
MS. Hence, a tangent touches the circumference in one
point only.
A secant is a straight line which, if produced, intersects
the circumference in two points, 'as GH.
yis. 3 Fis. *
202, A segment of a circle is a portion of the circle
bounded by an arc and its chord, as ABO (Fig. 3).
Into how many segments does each chord divide a circle?
A semicircle is a segment bounded by a semicircumfer-
enee and its diameter.
203. A sector of a circle is a portion of a circle bounded
by two radii and the are included by them, aa FOQ
(Fis, 3),
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THE cmcLE l05
204. A central angle is an an^lo whose vertex is at the
center and wliose sides are radii, as the angle POQ
(Fig. 3).
An inscribed angle is an angle whose vertex is in the
circumference and whose sides are chords, as the angle A/JG'
(Fig. 4).
An angle inscribed in a segment is an angle whose ver-
tex is in the are of the segment and whose sides are chords
drawn ft'oni the vertex to the extremities of tlie are. Let
the pupil draw a circle, a segment io it, and an angle In-
scribed in tlie isi'gnieut.
205. Two circles tangent to each other are circles which
are tangent to the same straiglit line at the same point.
They are tangent hifernally or externally according as one
circle lies entirely within or entirely without the other.
See the figures, page 122.
Concentric circles are circles which have the same center.
Let tlie' pupil draw a pair of concentric circles.
206. A polygon inscribed in a circle is a polygon all of
whose vertices lie in the circumference of the circle, as
ABODE (Fig. 4).
A circle circumscribed about a polygon is a cii'ele whose
circumference passes through every vertex of the polygon.
207. A polygon circumscribed abont a circle is a polygon
all of wliose sides are tangent to the circle, as PQlttiT
(Fig. 5).
A circle inscribed in a polygon is a circle to which all
the sides of the polygon are tangent.
Coacyclic points arc poiuts lying on the same circum-
ference.
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1U6 BOOK n. I'LAKE GEOMETRV
PROPERTIES OF THE CIRCLE INFEERED IMMEDIATEIY
208. liiidii of lli<: same circle, or of equal circles, are
equal.
209. The flianicier of a circle equals fu-ice its radius.
210. DhiDiefera of the same circle, or of equal circles,
are equal.
211. If two circles arc equal (i. e., ma}/ be made lo
coincide. Art. 13), their radii are equal, and conversely.
212. A diameter of a circle bisects the circle. For, by
placing the two parts of the circle so that the diameters
coincide and their arcs fait on the same side of the diame-
ter, these arcs will coincide (Art. 197).
213. A straight line cannot intersect a circle in more
than tivo points. For, if a straight Hue can intersect a
circle in three (or more) points, three or more equal lines
(radii) can be drawn from the same point (the center) to
the straight line. But this is impossible {Art. 111).
Proposition I. Theorem
214. -4 diameter of a circle is longer than any other
chord.
Given AB a diameter, and CD any other chord in the
circle 0.
To prove AB > CD.
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THE CIRCLE
107
Proof. Draw Uit radii OC and O'D.
Then, in the A OCD, OG+OD> CD. (Whyf)
Substituting for OC its equal OA, and for 01} its oqud
OB. Ax. s.
0A + Oh, or Ali> CD,
Q. E. B,
ri!Oi'OfilTIOX If. Tbeoukm
215. 7/i tiie same circle, or in equal circles, equal cen-
tral amjhs i)i/crcepl cijuul arcs oil the circumference.
Given the equal eircios and 0', aud tlie equal central
A. AOBaviAA'O'B'.
To prove the are AB=are A'B' .
Proof. Apply the circle 0' to the circle so that the
center 0' coincides with the center 0, and the radius O'A'
?.'ith the radius OA.
Then the radius O'B' will fail on the radius OB,
iforhjhi/p. lAOB^lA'O'B'j.
And B' will fall on B, Art. 20fi.
(/or OB = 0'B', being ra<Ui o/i-<pi<i! ®).
Hence arc A'B' will coincide with AB, Art. I!i7.
(/or uU poiiils of each arc arc iijiiUUslanI from (JiC i-'cnffr).
.'. are A'B'= arc AB. Art. 4T.
Q. £. D.
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108 l;OOK n. I'LANE UEOME'IliV
PrOTOSITION III. ThEOEEM (CONVEE5G OF pROP II)
216. In the same drch, or in eguai circles, equal crn-n
subtend equal augks at the center.
Given the equal (;irek's niul 0', niiil the efjual ares
AB and A'B' subteiuliug the central A ami 0'.
To prove ZO = ZO'.
Proof, Apply the eirele 0' to the equal circle so that
the center & colncii^es with the center and the point A'
with the point A.
Then the point B' will fall on B,
(for arc A'B'^arc AB iy hyp.).
Hence the radius ffA' will coincide with OA, and radius
O'B' with OB, Avi. CA.
{betuieeii two points only one straight line can he liAiini).
.'. L and £ 0' coincide.
.-. 10= 10'. Art. 47.
0, E. D.
217- Cor. In the fame circle, or in equal circles, of
two unequal central angles the greater angle intercepts the
greater arc, and, conversely, of two unequal ares, the greater
arc subtends the greater angle at the center.
Bx, Draw a, circle and hi it a segment whicb la less tbau tlie
sector hiviug the same arc. Also oue tiiat is greater. A\ea one that
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THE CIliCLE 109
Proposition IV. Theorem
218. In fhe snme. circle, or hi eqmil circles, equal chords
subtend equal ai-cs.
Given the equal eirelos; and 0', uud the elioi'd AB =
chord A'B'.
To prove are AB = arc A'B'.
Proof. Draw the radii OA, OB, O'A', O'B'.
Theu, in the A AOB and A'O'B',
AB^A'B'. (Wbr I)
AO=A'0', and BO=^!i'0'. (Why?)
.-. A AOB^AA'O'li'. (Why?)
.-. Z0= 10'. (Why J)
e O, .
;, arc AB = ai-c A'B'. Ari
y fv/iio; ®, eqiKil cmibut A intercept equal a
215.
Ex. 1. I„ the above f.^'ure, if thord AB=l iu., chord .
XB'--=.
aDaare^B = Uin., find the lenRth of iire .J'JJ'.
Ex. 2, Draw a clrtlt aud mark off a part of it that is
both V.
ment HDd a seolor.
El. 3. If the diatance fvoiu the eentor of a circle t
.0 a li
greater than the radius, will tha line iiitetWHct tUo ciri:HHi
fercuui
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10 UOOK II. PLAMi GEOMETllY
Piiopoarnux V. TheoiU':m
219, III ill'' s"»f'' rlrrlr, '/r in equal circles, equal arcs
>■€ suhlcnded h>j equal eJiurds.
Given the. equal circles O and (T, amlara AB^iiva A' B'.
To prove oliord AD — eliord .1'/!'.
Proof. Draw the radii AO, BO, A'O', B'O'.
Tlien, ill tlie A AOB and A'O'B',
10 = Iff, Art. 216.
(for are AJi = A' B' , iiiid, in Ilia snmo O, or in equal ®, equal a
U-ml equal A at ihe center).
Also OA = 0'A', and OB^O'B'.
(Why ?)
.-. A AOB- A A'O'B'.
(Why ?)
.-. AB^A'B'.
(Why f]
A'J!
Q. E. B.
Ex. 1. In the above figure if ara 4B = 11 in,, arc
ord Ali^l iu., find chord A'B' without meaauring
' = ljlii., and
Ex. 2. Draw two circles so that the radius of oc
the other.
. i.
the diameter
Ex. 3. To which o£ the elassea of figures mentiousd iu Art. 16
oes a sector ijelong f a segment i
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the circle 111
Proposition VI. Theorem
220. In the same circle, or in equal circles, the greater
of iv:o {minor) arcs is subtended by the greater chord; and.
Conversely, the greater of two chords subtends the greater
(minor) arc.
Given the equal circles and O', and are AB > aro 1>F.
To prove chord AB > chord DF,
Proof. Draw the radii OA, OB, O'D, O'F.
Then, in the A AOB and DO'F,
OA=0'D, and OB^O'F. (Why?)
Z is greater than Z 0', Art, 217.
(for arc, AB > arc I)F, and, in the same G, or in equal 0, of two un-
equal arcs the greater are subtends lite greater angle at the center) .
.■. chord AB > chord DF, Art. lOT.
(if !«■(> A have two sides of one eqitat to tiro siiles of the olhci-, but the
included angle of the first greater, etc.).
CoNVKRSELY. Given the eqiial eh-eles and (y , and
chord AB > chord DP,
To prove arc AB > are DF.
Proof. In the A AOB and DO'F,
OA = 0'D, and OB^O'F. (Why!)
AB> BF. (Why?)
.•. Z is gi-eater tlian Z 0'. (wiiy ?)
.-. arc AB > are BF, Art. 2IT.
(in the name O, or in ei/Ma! ®, of tiro viirqual central i llie grimier
"™"" '"^"' e, £. B.
tnyle intercepts the greater
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112
liOOK a. I'LASE (iKOME'lMlY
Proposition VII. TiiEOREJi
221. ,1 iiiameter perpendiculnr to a chord hi-wcls the
cJwrd and the arcs subtended by the chord.
Given tbe circle 0, and the diameter PQ X ch.
To prove that PQ bisects the chord AB and the ai
and -1 <JB.
Proof. Draw the radii 0-4, and OB.
Then, in tlic rt. A OAR and OBR,
OA = OB.
OB = OB.
:. AOAB = A OBB.
:. AE=BR. and ZAOB = I BOB.
:. are .1 P. = arc HP,
(in ihr .wmr O, n;- in = ©, = central A intercept =
i-d AR.
isAPB
(Why?)
(Why?)
(Why?)
(Why!)
Art. 215.
■<■/.)■
^AOQ ::=: ^ BOQ {-'^A 15). .\ HTQ AQ ^ ATC BQ. (Wbj t)
Q.E. D.
222. Con. 1. -i diameter which l/isecis a chord {shorter
than a diameter) i« perpendicular to the chord.
223. Cor. 2. The perpendicular bisector of a chord
passes through the center of the circle, and bisects the arcs
subtended hy the chord.
224. Cor. '6. A line from the center perpendicular to
a chord bisects the chord.
225. Cor. 4. A line passing through the midpoints of
a chord and Us arc passes through the center, and is a diame •
ter perpendicular to the chord.
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THE CIRCLE 113
Proposition VIII. Theorem
226. Ill the same circle, or in equal circles, equal chords
are eqiudisiaitl from the center; and, conversely, chords
which are equidisiani from the center _ are equal.
Givea the cii-ele in wliich the eliords AB ami CD
are equal.
To prove that AB and CT) tire eqnidistant from the
center.
Proof. Let OE he drawn ± AB, and Of ± CD.
Draw the radii OA nnd Of.
Then OE bisects AB, and OF Wsects CD, Art. 224.
(a Jiiicfriym the milcr ± n chord biser.ls the chord).
Hence, in the rt. A OAE and OCF,
OA = 00. (Why?)
AE = GF. Ak. s.
.-. AOAE ^ A OCF. (wiiy?i
.-, OE = OF. (Whj?)
Conversely. Given circle 0, and AB and CD equidis-
tant from tlie (tenter.
To prove AB=CD.
Proof. Let the pupil supply thu proof.
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114 POOK II. l'LA>JE GEOMETRY
Proposition IX, Theorem
227. In the same circle, or in equal circles, if two chords
are unequal, they are uHequally distant from the center, and
the less chord is at the greater distance from the center.
Given in the circle the chord CD < chord AB.
To prove that chord CD is at a greater distance from the
center than chord AB.
Proof. Let OG be drawn 1 CD, and OF X AB.
Then chord AB > chord CD. Hyp.
.-. arc AB > arc CD, Art. 220,
(iH the sameO, or in eijml. ®, Ihe greater of two minor arcs is svbtendiil
h}i Ihe ijrcat'.T clwrd, mid coiwcrscly) .
Mark off on the arc AB the arc AE^&n: CD, and draw
the chord AE.
Chord AE = chord CD, Art. 219.
(t» the same O, of in equal ® , equal arcs are subt/ntiled by equal cltords).
Let OS he drawn ± AE, and intersecting AB at L.
.: OS = OG, Art. 226.
(in the same O, or in equal ®, equal chords are equidistant from the
But 0S> OL. Ai. 7.
Also 0L> OF. (Why!]
Much naore then OS, or its equal OG > OF. Ax. 12.
Q. B D.
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THE CIRCLE 115
pRorosiTiON X. Theoeem (Converse op Peop. IX)
228. Jn the same circle., or in equal circles, if iwo chords
are nneqimlly distant from the center, the more remote is
the less.
Given in t!iR ciriilc llie olioi-d VI> farther from the
center thao tlio chord All.
To prove ehunl CD < cJioiil J /.'.
Proof. Let OH be (iiwwn 1 CTi, mul OG ± AB.
OH > CXr. (Why?)
Oil OH mark off <>L = 00.
Through L let the eliord ELF he drawn X OE.
Then chord EF = chord AB, Art. 22G.
(in Hie same ,C
But the are CD < are EF. As. 7.
.-. chord CD < ehord EF, or its equal AB, Art. 220.
(in (fte same Q, or i« fqnn/ ®, ilie greater of tico minor arcs is sublended
by ilic greater chord, and conversely-),
Q. £. 9.
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lit; BOOK IT. TLAXE GE03IETK1'
PliOI'OSlTION XI. TUEOKEJI
229. A KiraUjht line pcrfeiuUcular to a nidiits at Us
fxtreinitii is Unnjint to tiic circle.
Given the circle 0, tlie i-adiua OA, and \hc line HC ± OA
at its extremity A.
To prove that BC is tangent to the .
Proof. Take P, any point on the line BC except A,
and draw OP.
Then 0/' > OA. (Why!)
Hence the point P lies witliont llie circle.
.'. every point in the line BC, except A, lies ontside
the O.
,■. BGig, tangent to the circlf, Ari. 2Ci.
(« (ann^^il toaO is a straight hi,,- ichkk, etc.),
0. E. B,
230. Cor. 1. The radius drami to the point of contact
is perpendicular io a tangent to a circle.
231. Cor, 2. A perpendicular to a tangent at the point
of cantad passes through the center of the circle.
232. Cor, 3. The ■perpendicular drawn from the center
of a circle to a tungent passes through (he point of contact.
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THE CIRCLE 117
pROPOSiTioK xn, Theorem
233. Two parallel lines intercept equal arcs on a eir-
Case I. Given AB (Fi<r. 1) fcangeat to tlie G I'CD at
P, CD it sGiJiuit II AB iiiid iuterseGtiug the circumfciuiieo in C
andi>.
To prove arc PC = are PD.
Proof. Draw the diameter PQ.
Then PQ ± AB, AH. 230.
PQ 1. CD. Art. 123.
.-. are PC = are PD, Art. 221.
(a diameter X chord iisR./s (/ib diw.f iKiiJ the ai-e« stMendea by
the chord).
Case II. Given .iB and CD {Fig. 2) [| seciiiits inter-
secting the circumference in A, B and C, D respeefcivoly,
Tb prove arc AQ — are BD.
Proof. Lota tiLiigent ^i^ he drawn |UIB unLltouehing tlie
circle at P.
Then EF\\ CD. (Wliy?)
Then arc AP = arc BP. Case I,
:^ CP= arc DP, CVVhy!)
Also
Hence
arc AC -
iBIJ.
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118 IKJOK II. rLAXK CEOMETRY
Case III. Given Ali and CI) (Fig. 3) \\ t;uigeiits toucli.
iag the O at P and Q respectively.
To prove arc PEQ = are I'FQ.
Proof. Let the tmpil supply the proof.
234. Cor. The straight line whk-li joins the points of
contact of two 2}ctraUel tangmiix is a dinmeier.
Peoposition XTIT. Theorem
285. Through three points, not in the sa
%e circumference, and only one, can be ih-a
Given A, B am! C any three points not in the same
straight line.
To prove that one eirenmference, and only one, can be
drawn through A, B and C.
Proof. Draw the straight lines .4^ and BC, and let _b
be erected at the midpoints, D and E, of AB and BG
respectively.
These -li will intersect at some point 0, Aft. vi'i.
{lines ± nmi-paralUl lines are not || ).
But is in the X bisector of AB. Const.
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THE CIRCI,E 119
.*. is equidistant from the points A aud B. Act. H3.
In like manner, is in tlio J. bisector of BC, and is
equidistant from the points B and C.
Hence ia equidistant from the three points A, B and G
Henee if a circumference be described with as a cen-
ter and OA as a radius, it will pass through .4., B and 0.
Also DO and EO intersect in but one point. Art. e4.
Hence there is but one center.
Again, is equally distant from the points A, B and C;
henee there is but ojte radius.
With only one center and only one nuliua, but one (lir-
cumference can be described.
Hence one circumference, and only one, can be drawn
through the points A, B and C.
Q. E. B.
236. Note. The theorem of Art. 235 enables ua to shrink or
economize a circle into three poiiits ; or to expand any three points
into a circle.
Ex. 1. How many eircurafereneea can be passed through fouf
given points in a pliine, each cireumferenee passing tbrou;i:h three,
and only three, of the given points I
Ex. 2. Draw two eirclea so that they outi have a common chord,
Ex. 3. Can two circles whii-h are tangent to each other have a.
common chord !
Ex. 4, Can two circles which are tangent to each other have s,
common secant 1
Ex. 5. Draw two circles which can have neither a
nor a common tangent.
Ex. 6. Is it poBsibie to draw two civulus which
common secant f
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I'LANE GKOMET:
Proposition XIV. Theorem
237. The two iangenis drawn io a circle from
outside the. circle are cqiinl, a>id make, equal (o/y/cs
line drawn from the point to the center.
point
icith a
Given PA and PB two tangents drawn from the point
P to the circle 0.
To prove PA^PB, and ZAPO= ZBPO.
Proof. Let the pupil supply the proof.
238. Dbf. The line of centers of two circles is the
line joining their eentei's,
239. DBF. Two circles which do not m(!ct m»y have
four common tangents.
A commoa internal taageat of two circle.'^ is a tangent
which cuts their line of centers.
240. Dep. a common external tangent of iwo circles is
3. tangent which does not cut their line of centers.
Kx. 1, In the above Sgure, prove that the line ilrawu to the center
from the point in which the two tangents meet makes equal aiiglsB
with the radii to the points of eontaet.
Ex. 2. Draw a circle with a radius of 3 ia. and anothei' with a radius
of 3 in., with their oeutera 4 in. apart. "Will these eirelea intersect T
n their centers were 6 in. apart, would they intersect S
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THF, CIRCLE l^J
Proposition XV. Theorem
241. Iftico circles intersect, their line of centers is per-
pendicitlaf to their common chord at its middle point.
Given the circles ami 0', itilevscetinK at tlm points
L and B.
To prove 00' ± AB at its middle point.
Proof, Draw the radii OA, OB, O'A, O'B.
Theu OA^OB, und 0'A = 0'B. (Why?)
Hence and 0' are two points each equidistant fi'oin
1 and B.
:. 00' is tlie ± bisector of AB. ah. in.
Braw two cirules in which tlie Una of e
Ex. 2. Equals the sum of the lailii.
L and />.
.-. 00' is tlic X bisector of AB. Avt. 113.
Q. E. D.
Draw two eireles in which tha 1
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V22 liOOK ir. I'LANK (lEOMETKY
PiiOPOhiiTioN XVI. Theorem
242. Jf two circles are iaiujeut to each oiker, fke Uite of
eenlers passes (hrough the point of contact.
Given tho cirdes iiml 0' tangfiiit to each otliev at tbe
poiut R.
To prove that the line 00' passes through It.
Proof. At the point E let PQ, a tangent to the given
®, be drawn.
Also let AB be drawn 1 PQ at R.
Then AB passes through and also throngh 0', Art. 231,
(a l-loa tangeni al the fioint of coiilact pusses Ihrough the center).
:. line AB eoincides with the line 00'. Art. 64.
.'. 00' passes through the point E,
(Jor it coiittidm ititli AB ahieh passes through li ) .
How many eomnioil interail, and
tangents bave two eirolea
Ex. 1. If they toneh esternally t
Ex. 2. If they touch iDternally f
E*. 3. It they intersect !
Ex. 4. If one circle lies wholly with
Ei. 5, If one circle lies wholly witl
external
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THE CIRCLE
eXERCISES. CROUP I
Ex. 1. The line joining the oenlor
cKord is perpeadienlar to the chord.
I four points t alt en i
'O is tho perpendieul
1' upon two clioi'ds a
[■0 BD; prove that chord
Ex. 2. A, B, Cand D n.
eircumferenee of a eirele,
AC=ahovii BD.
Ex. 3. Taugeuts drawn at the estron
p&ratlel.
Ei. 4. PA and FB aro tangents to a eii
P. is the center of the oirele. Prove th;
bisector o£ the chord AB.
Ei. 5, If the perpendiculars from the Ci
equal, tlio aroB subtended by these chords a
Kl. 6. A, B. C and D are points taken i
onmfecence, and arc AC is greatei' than t
AB> chord CD.
Ex. 7. State the converse of the preceding
theorem and prove it.
Ex. 8. Given EA, EQ and QB tangents o£
the circle ; prove Rq=EA + QB.
Ex. 9. If a quadrilateral be e ire urns cribed
about a. eirele, show that the eum of one pair of
opposite aides equals the auiu of the other pair.
Ex. 10. if a hexagon be eireumscribedabout aairolo, show that tho
aiim of three alternate aides equals the sum oC the other throe sides.
Ex.11. If a polygon of 2rt sides
(he sum of n alternate sides equals t:
Ex. 12 Acm
inequilateral,
Ex. 13, If two circles b
externally, the eoiuiuuii intei
bisects the common external ti
prove PA = FB).
ribed paralleiograi
mgent
.tl,i.,
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il;4
BOOK II. I'L.VNR nKOMF.TliV
Ex. 14. It two circles are tniigfi
(either e.tteniiilly or iiitei-nally) taugcti
drawn to llieiu from any point iu tie eon
mon tangent ara equnl.
Ex. 15. Two eircli
O' are tangent mternnJly
at P. The line PAB ia
drawn intersecting the
Prove that OA and O'B
MEASUREMENT. RATIO.
243. Measurement. For many purposes, the most;
advantageous way of ilealiog with a given magnitude is to
take a certain definite part of the magnitude as a unit,
and to determine the nnmbei' of times this unit must be
taken in order to make up the given magnitude. Ease and
precision iu dealing with magnitudes are thus obtained.
Geometric magnitudes thus far have been treated aa wholes, the
object being simply to determine whether two given magnitudes are
equal, or unequal, or to determine some similar general relation.
Hereafter geometric magnitudes will frequently be treated as if com-
posed of units.
To measure a given magnitude is to iind how many times
the given magnitude contains another magnitude of tlie same
kind taken as a unit.
244. Tlie numerical measure of a magtiituJe is the num-
ber which expresses how many times the unit of measure is
contained in the given magnitude.
Thns, when a boy says that he is five feet tall, he means that, if afoot
rule be applieti to Lis height, the foot ruie will be contained live times.
A quantity is often measured to beat advautage by measuring a
related but more accessible quantity, which Las the same numerical meas-
ure as the original quantity. This process is indirect measurement. Thus,
the temperature of the air is measured iniiirettly by measuring the
height of a column of mercury in a thermometer tube. So the number of
times a unit of angle ia contained in a given angle is often ascertained
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MEASUKKMKKT
125
moat readily by determining tbe number of times a unit of arc is cod-
taiued ttt a give □ a. re.
245. The ratio of two niiigniUKius of the same kind is
their relative magiiitatle as determined by the number of
times one is contaioed in tbe other. Hence, it is the qiio-
tient, or indicated quotient, of tlie two miignitudos.
Thus, tbe ratio of 3 ft. to I ft. 7 in. ia ^~. <"■ fg-
Tlie use of ratio U illustrated by tUe fuct tbat several indicated quo-
tiimts wbcil taken togetlier may be simplified by canoellation before a
floal determination of tbeir vnlins La iiinde.
Two magnitudes of the same kind have tlie same ratio as
their numerical measures.
246. Measurement as a ratio. Au important piirticular
instance of ratio is tbat ratio in wliicb one of tbe two mag-
nitudes compared is a unit of measurement. Hence, the
nnmerical measure of a magnitude is the ratio of tbe mag-
nitude to tlie unit of measure.
Tims, tlie numerical meaaure of tbe beiglit of a boy is tbe ratio of,
bisIieigU(5 ft.) to tbe uait of measiiru (I ft.), orS.
247. Commensurable maguitudes are magnitudes of the
same kiud which have a common unit of measure,
Tiius, 12 ft. and 25 ft. bave tbe common unit of measure, 1 ft., and
ience are commensurable magnitudes ; also 13! bu3, and 7J biia, liava
a common unit, I peclc, and are commensurable,
248. Incommensurable magnitudes are magnitudes of
the same kind wliich have no common unit of measure.
Ills, $5 and Iv'a ; 7 vrs. and -^15 yrs.; so tbe side and the diagonal
of a squHve may be proved to bave no common unit of measure ; likewise
tbe diameter and tlie circumference of a circle.
In general, a ratio wliicb is expressed bj a surd number,
^\/S or\ 3, is a ratio between incommensurable magnitudes
(called au incommensurable ratio).
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'I,ANE GEOMETRY
METHOD OF LIMITS
a variable, is fi qnanlity
imlK'f of differfijit valnes
or |3voi)ltMii .
goes varies with the nurabei of
tou w
h
250
\ constant
u
ivhicli remains iincliiiiigeLl
in \
u a
Lissioii.
h
g
circle and tlie iiuinUcr of sides
oftb
itc, the pariiiietet of tUe pnly-
j;OI\
e 1)
iioe of tlie circle nill remain
uncLan^
an
li b
u
251
T
mto
JJ
mtity is a conatiiut quantity
^fiven discussion, the given
\-M
5 we please in Talue, but
1 the circumference of tlio circle
]a til lie n e bed i>olygun ; also,- the area of
fUe circle is the limit of the urea of the inscribed polygon.
From the definition of limit, it follows that the differ-
finee between a variablo and its limit may be made as small
as we please but can not become zero.
As another illustration of a variable and its limit, we may talie the
ease of a point Ptravelling along a given line AB, in auch ft way that
in the first second it passes over APi, one-half the line AB; in the
second second over half the remaining part o£ the line, and arrives it
P;] in tlia third second over one-
half the remainder of the line, and , , , ,
arrives at Pj, etc. It is evident -^ fi ^' ^^ ^
that the point P can never arrive at
B; for, in order to do this, in some one second the point would need
to pass over the whole of the remaining dietanee.
In this illustration, AP, the distance traveled by the moving point, is
a variable (depending on the number of aeconds), and AB is its limit.
If the distance AB be denoted by 2, the distance traveled will be
denoted by 1 -fl + i -|- i + , . . . and will vary according to the
number of terma o! the series taken.
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SfETHOD OF LIMIT3
127
252. Use of variables aad limits. Many of the proper-
ties of limits are the same as tlie properties of the variables
approaching them. Ileuce a demonstration of a difticiilfc
theorem may often bo obtained by first fintling the properties
of relatively simple variables and then transferring these
properties to their more complex limits.
253. Properties of variables and limits.
1. The liinit of iJte sum of a nitmbfr of eari^lfS equals
the sum of the limits of these vaHables. For, since the
difference between each variable and its liinit may be made
as small as we please, the sum of alt these differences may
be made as small as we please (since it is a finite number
of differences, with each difference approaching zero).
2. The litnit of a times a variable equals a iiiiies the
limit of the variable, a being a constant. For, if the differ-
ence between a variable and its limit may be made as sinall
as we please, a times this difference may be made as small
as we please.
3. The limit of -tk part of a variable *■'< -fh part of the
limit of the variable, a being a constant. For, if the
difference between a viiriable and its limit may be made an
small as we please, — th part of tliis difference may be made
aa email as we please.
4. If a variable ~0, a times (a being finite) or - part of
the variable ~0; and if A diminish in the one, or i
the other process, the limit is sliU zero.
Ez, 1. Are 2) gal, mid 3^ qt. c
Ei. 2. If c donotc u miistaiit bi
eacU ot the following a v&riable or a
Uftte.
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LASE (iEOJlETRY
XVII. Tin:f)iu.:M
254. 7/ tu:> rariabU^ nr- .iJmnj^ pqml, and mch <
.■or.clu^.i a UntU, /luir Umil^ un- 'qiuil.
y I> <-
GiTen AB flic limit of the variable Al', AC the Ihuit of
the variable AQ, and AP^AQ always.
To prove AB = At7.
Proof. If the limit AB does not equal the limit AC,
one of these limits, as AG, must be larger tiiau the other.
Then, on AC, take AD equal to AJi.
But AQ may have a value greater than AD. Art. U.jl,
Hence AQ would be greater than .i/>,
Or AQ > AD tt-hi.)h = AB) .
. AQ > AP,
{Jor
AP).
But this is eontrary to the hypothesis that AQ aud AP
are always equal.
,'. AC cannot be greater than AB.
Id like manner it may be shown that AB is not greater
than AC
:. AB^AC.
Ex, "Wliiit method of proof is used in Prop, XVII!
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MEASUREMENT OP ANGLES
Proposition XVIII. Theorem
255. In the same circle, or in equal circles, ix-o
angles have the same ratio as their intercepted arcs.
Case I. T17jfw the intercepted a
tral
'mmensuraile.
To prove
Given thuoqiuil® C and 0, with tlie ceutrai ^A'O'B' ami
AOB intercepting the commensurable arcs A'B' and AB.
ZA'O'B' ^A'B'
lAOB AB'
Proof. Let arc PQ {from n circle - ami 0') be ix com-
mon moasiiro o£ tho arcs A'B' and AB.
PQ will be contained in arc A'B' an exant number of
times, as 7 times, and in AB an exact number of times,
as 5 times.
arc AB 5
From (y and draw radii to the several points of
division of the arcs A'B' and AB.
Then the ZA'O'B' will be divided into 7, and the
/.AOB into 5 small angles, all equal. Art. 216.
{in the «!.
Fj =® equal arcs mbtend eqviU A. at the t
lA'O'B' ^1
lAOB 5
ZA'O'B' ^A'Ii'
I AOB AB'
Art. 245.
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BOOR II. I'LAXE
Cask II. inint the h<lr>rrpf''d an
siirahle.
Giventheeriiiii! ciiO'iimlO,witli the eeiiti'id A A'O'B' a.hA
AOB iuteruepting the iuooramensorable arus A'B' and AJi.
To prove
ZA'O'B' _
' lAOli
AlV
AH '
Proof. Let tlie arc AB be divided into any naniber of
equal parts, and let ooe of these parts be applied to the
arc A'B'. It will be contained in A'B' a certain number
of times, with an arc BB' as a remainder.
Henoe the arcs A'D and AB have a common unit of mess-
lA'O'D A'D
lAOl) AB '
Case I.
If now wu let tlie unit of measure be iiidcflii
ilelydimin-
ished, the arc DB', which is less thun tlie unit of i
ncasuro.will
be indefinitely diminished.
Hence arc ^'i>=^arc A'B' a-s a limit, and
I A'O'D-
^1^'0'5'asalimit.
An 351.
Hence , . ..„ becomes a vafiable approaohi
ZA'O'B'
" lAOB
as its limit;
Alt. 253,3.
Also -j^ becomes a variable approaching
A'B' .
limit.
Art. 253, 8.
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MKASri!EMENT OF ANGLER 131
Lsel.
256. Dep. a degree of arc is one tliree luiiiJf.'d and
sixtieth part of the eircura fere nee of h (circle.
257, Cor. TJie, mcniber of degrees in a cfnttul umjle
equals the niimher of degrees in /he hitercepled an: ; thut is,
a central angle is measiirrd by Ua Intercep/ed are.
Ex. 1. What is the ratio ot a quadrant to a semi-o Ira u inference !
Ex. 2. "What is the ratio of an angle of an equilateral triangle to
ne of the acute angles of an isoaeeles riglit triangle t
ot each circle is on the
Ex. 4. Draw three circles so that the center o£ each ia on
camterenee of the other two.
[SuQ. First draw an equilateral triangle.]
Ek. B. Draw three circles each of which shall he tangei
other two.
Ex. 6, Draw tv
to one ot these cirt
Ex. 7. Draw two concent rit
one and a chord o£ the other.
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liOOK II. PLANE GEOME
Proposition XIX. Tbeork.u
258. An inscribed angle, is measured by one half ifg
Inlercepted arc-
ana AB
(Whyf
(Whyfl
Case I, When the n-nl
the inscribed angle.
Given Z2J.4C(Fig. 1) iusei-ibecl iii the I'ircle C
passing through the center 0.
To prove that Z BAC is measured by * aw. BC
Proof. Draw the radius OC.
Then, in the A OAC, OA = OC.
I A = IG.
But lIiOC= ^A +Z0.
.-. ZB0C = 2Z.-1.
But Z HOC is measured by arc RC,
{aceitlral Z iiineo-mredhmlsixlcrccincd arc).
.'. Ko. of iingulur degrees in / .BOC^No. of arc dfgi
Hence Z -i is iiieusnred by \ arc SC. Ai. 5.
Case II. Wheii the center of the circle lies within the
ht scribed tmgle.
Given the inscribed /IBAG (Fig. 2), with the eentec
of the circle tying within the angle.
To prove that /.BAC is measured by h arc BO.
Proof. Draw the diameter AD,
Then /.BAD is measured by ^ are BD. Caael.
Let the pupii complete the proof.
aBC.
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MEASUREMENT OF ANGLES
133
Case III. When (he cenier of the circle is witliout the
inscribed angle.
Given the inscribed IBAG (Pig. 3) with tlie center
outside the angle.
To prove that ABAC is measured by h arc BC.
Proof. Let the pupil supply the proof,
259. Note. By use of the nbove theorem, if the number of
degree* in the intereepfed arc be known, the number of degrees in
the insoribed Bnslo enn be determined immediately. Thus, if the arc
BC (Pig. 3) eoQtaixiS 4S°, the angle BAC contama 24°; also, if it be
known that the mslo BJC e<
260. CoE. 1. AU angles inscribed in the same segment,
or in equal scgmeni.t, are equal.
Thns A A, A', A" (Fig. 4) are all equal, for each ot
them is measured by one -half the arc BBC.
261. Cob. 2. An angle inscribed in a semicircle is a
right angle, for it is measured by one-half a semicircum-
ferenee.
Thus FE (Fig. 5) is a diameter .-. ZFGEis a rt. Z.
262. Cor. 3. ^1k angle inscribed in a segment greater
fhan a semicircle is an acute angle; an angle inscribed in a
Segment Jess than a semicirch is an ohttise angle.
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iS-l BOOK II. PLANE CEOMI'mji:
Proposition XX. Theoreji
263. An angle formed bi/ two diordn Intersecting leiihin
a cireiimfiirence is medsurrd hij half the sum of the inter-
cepted arcs.
Given tlic chords AB and CD in the G AVBC, inter-
secting within the circiimferenco at''tlie point P.
To prove that /.DFB is measured by J (arc BB-\- ai-o
AC).
Proof. Draw the i^hord AB.
Then, in A ABB,
IDBB= IA+ IT), Art. 13S.
[an exi. I of a £\ is equal lo the sum of the lu'o opp. int. i ).
But Z J. is measured by i are DB, Art. 358.
{an inscribed Z is measured l/ij one-half its mlcrvepte/i are).
Also Z 7> is measured by i are AC. (Why!)
Hence I DFB is measured by A (arc JJB + arc AC).
Ax. 3.
Q- E. B.
Ex. 1. lu the above figure, i! arc />/( eonUins W and are -IC coa-
t*ms 38°, how many degrees in lAPC * ia lAPD ?
Eic. 2. If arc AD = 5i;° and Z AFD = 124°, fiud are CB.
Ex, 3. If, in Fig. I, p. 132, arc AC oofttalna 112°, how many
degrees are there in tiie I A !
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MEASUREMENT OF ANGLES 135
Pedposition XXI. Theorem
264. An angle formed by a tangent and a chord drawn
from the point of contact is mfamred by half the inter'
cepied arc.
Given the O PCD, awl Z.APC formed by the tangent
APB and the chord PC.
To prove that Z APC is measured by i are PEC.
Proof. Let the ehord CD be drawn ll AB.
Thtii
Z.APC -^
■- I PCD.
(Why?)
But
I PCI) is maasu
red by I a
.i-c PD.
(Why!)
.-. I APC is measu
red by i a
re PD.
Ax. 8.
But
(i.
arc PEC =
00 W Uiie,^ inkrcept cijuw
arc DP,
■rcumfermce).
Art. 233.
,■, /APC is measured by i arc PEC. Ax, 8,
Also Z.BPC, the supplenieut of ^APC, is measured
by i are PDC, Ax. 3.
(for urc I-DC is the conjmatc of arc I'liC) .
Q. E, B.
Ex. 1. In the iibove figure, if are /'A'(,* uoKtaius l::4'^, Low many
degrBea lire in the angle AI'C !
Ei. 2. If aru CD = 'M", find the angles on the figure.
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rKorosmoN XXJT. Thkorrjc
265. An an'jle formed by two secants, or hij lirn tangents,
or hi! a secant ami a tangent vieel in;j withmil the circumference
is measured by half the ilifferntre nf Hip hilerrrjiled ores.
FiS.t 3?is.%. (fib', a
I. Given the O ACDB (Fi^. 1), and the lAPB formed
hy the two secants PA and P£, meeting at the point P with-
out the circumference.
To prove that ZP is raeasnrerl hy 1 (arc>4B— arc CD).
Proof. Draw tlie chord CB.
Then IACB= ZP+ZB- [Whj-i)
.■. ZP= ZACB— ZB. Ax.3,
But ZACB is measured by i arc AB. Art. 258.
Also IB is measured by 4 arc CD. (Why!)
.•. /P is measured by A (arc ^P — arc CD) . Ax.s.
II. Given the O ABB (Fig. 2), and ZCPB formed
by the tangent PC and the secant PP.
To prove that Z P is measured by * (arc ABB — Are AD).
Proof. Let the pupil supply the proof.
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KXKliCTSF.S OK THE CTRCLK 137
m. Given Z^Pfi (Fig. 3) formed by tlie tangents PC
and PD.
To prove that ZP is measured hy i {arc AEB —
are AFB) .
Proof. Let tho i>iipil supply the proof.
266. Note. By means of Props. XVIII-XXII, angles formed
by chords, seeanta, or tangents, or oombinations of these, are all
reduced to eentrp.1 angles and hence may readily be compared.
Ex. 1. If, in Tig. 1, p, 130, are CI> = Zi° and arc .JJV = lOS",
draw AD and find all the angles of the figure.
Ex. 2. If, in Fii-. 2, p. 13U, angle i' = SO'', find arcs-Ji'fl and AEB.
EXERCISES. CROUP tS
Ex. 1. What now methods of proriug two lines equal are afforded
by Book ir t
Ex. 2. Wliat new methods of proving two angles equal '. of prov-
ing two angk-5 supplementary f of proving an angle a right angle 1
Ex, 3. What methods of proving two arcs equal !
Ex. 4. In ii quiidrilareral ioscribed in a circle, tiicli pair of uppiwiti;
angles is supplcinciitiiry.
Ex. 5. A «hord forms equal angle? with the tangents at ita
extremities ,
Ex. 6. If an isosceles triangle be inscribed in a circle, the tangent
at its vertex makes equal angles with two of its sides and is parallel
to the third side. (Is the converse of this theorem inic ?)
Ex. 7. I£ two chords in a circle intersect within the i^irele at right
angles, the sum of a pair of alternate area equals
a semioircumf erenee .
Ex. 8. Given the center of a circle and
-iC a tangent ; prove IBAC=X<<^0.
Ex. 9. If one side of an inseribed quadri-
lateral be produced, the exterior angle fio
formed eiiuals llio opposile interior hi
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i;;s
];ooK II. ]']^ANE ^.EO^rKTuy
Ex. 10. IE twi> tiingoiits (o a. oirele incluiio an nn^iia of 60" at
their point of intersection, the chord joining the ]ioiut3 of eontaot
forms with the tnngenta an equilateral tl'iangle.
Ex. 11. Tliu chord A/I equals the eijord C]) lu a given eirele, and
the chords, if prodneoii, iutei'svet at the point i'. Prove seeant PA^s
secant I'C
pribed ahout the triani^le ABC, and P ia
Prove that the angle Jill' equals oue-
Es. 12. A circle iE
the uiidpoint of the a
half the angle C.
Kk. 13. It A, n, C, 1> and E be poiTits taken ii
eireumtercnee of a eirele, and the area AU, BC, CIJ and /IE be equal,
prove that the angles AliC, HCI) and CDl! are equal.
Ex. 14. An inscribed angle formed by a diameter and a chord baa
its intercepted are biaeeted by a radius which ia ,,
parallel to the chord.
Ex. 15. Two see^mts, PAI! and PCD, intersect
the eirele ABDC. Prove that the trian>;les P/IC
and rj/>are mutnally equiangular.
Ei. 16. Given AC s, tangent ami An\\ri:,-
prove A ACD and AJIE rautnally equiangular. '
Ek. 17. Given ABC an inaoribed triangle, AE
± BC, and CD X AB ; prove arc iJO=arc BE.
Ex 18. If the diagonals of an iuseribed quad-
rilateral are diameters, the quadrilateral is a
Ex. 19. Tangents through the voitie.
inacribeii I'eptangle form a rhombus.
Ex. 20. Two circles intersect at /'
and Q. PA and Pli are diameters.
Prove that <^A and QD form a straight
line.
Ex. 21. Two eqi
at Pand Q, through P a line ia dra
terminated by the eireumfersooes
equals QB.
[SuQ. i A and B are measured by what arcs !]
J and B. Show that QA
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EXERCISES. AUXILIARY LIXEK
139
267. Use of auxiliary lines. In deraoustfatiii^ theorems
relating to tlie cifcie, it is often helpful to draw one or
more of the following; auxiliary Hues: A radius , a diame-
ter, a chord, a perpeirdicuhtr from the coikr upon a chord,
an arc, a circtimferfiu-i^, etc.
EXERCISES. CROUP 17
AC'XTI.IARY LIXKS
•J ; prove BM = r.X.
Ex. 2. If from any point in the eircuraferenee
o! a circle a eliord and a tangent be drawn, llic
perpendiculars drawn to tLem from the midpoint
of the are subtended by the chord are eq^ual.
Ex. 3. Given the center of a eirule, and 0J>
X chord AC ; prove IA0D= IB.
Ex. 4. Prom tbc extremity of a cHametcr chords
are drawn making «iual angles ivitli tiie diameter.
Prove that these chords are equal.
ISvG. Draw Ji from the center to the chords, et
la the converse of this theorem true ?
Ex. 5. If through any point within a circle equal chords he drawn,
show that the line drawn from (he center to the point of intersection
of the chords hisects their angle of intersection .
Ex. 6, Tangents /'J and W are drawn to a cir
O. Prove that angle i' equals twiue angle OAIl.
Ex. 7. If in B
the segments of o
other chord,
Ex. 8. A paru
rectangle,
CSuo.
P, 100.]
le Iwp equal chorda ir
orii equal the segment
( the diajjouiils o£ thu i—/ ,
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rr,ANF, riT'O.MKTRY
Ex. 9. If a qundvLlatera! Ije o ire am scribed nboiit a circle, Um anglea
subtended at the center by a pair of opposite aidfis are supplpinentacy,
[Sua, Draw radii to tlie poiiifce of contact and show that tiiere are
four pairs of equal A at Hie center.]
Ex. 10. If an equilateral triangle ABC be inscribed in a oirele and
any point P be taken in the arc AB, show that PC= PA+FB.
[SUG. On PC take Pif equal to FA, draw AM and prove ii PAR
and MJC equal.]
Ex. II. Two radii perpendicular to each otlier are produced to
intersect a tangent, and from the points of intersection othei' tangents
are drawn to the circle. Prove that the tangents last drawn arg
parallel .
[Sun. Draw radii to the three points of con- A
tact, etc.]
Ex. 12. A straight line intersects two oonot
trie circles. Show that the segments of the I.
intereepted between the eirpumferences are eqi
(prove --15 = 01)).
Ex. 13. A common tangent in
drawn to two circles wliich ara
exterior to each other. Show that
the chords drawn from the points
of fangenoy to the points where
the line of centers cuts the cir-
cumferences are parallel.
Ex. 14. A circle is described on the radius of a
diameter, and a chord of the larger circle is drawn from tlie point of
coataet of the two circles. Prove that this chord i
circumference of the smaller circle.
[SuG. If the chord is bisected, a X from the c(
the chord will also bisect ti:B chord, i
Ex. 15. Two circles are tangent ex-
ternally at the point P. Tiirough F any
two lines APB and CPD are drawn ter-
minated by the circumferences. Show
that the chords AC and BD are parallel.
[Sl'O. Draw the common tangent at
F, If AC and BD are |1 , what A mui
■ted hythe
of the larger to-
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EXERCISES. MAXIMA AND MINIMA
Ex. 16. Two circles intersert at the points P Biid Q.
and CQU uro drawn, terminated by the oiroumferences.
^Cand BDare parallel.
Ex. 17. If a square be described on the hypotenui
triangle, a line drawn from the center of the square to
the right angle bisects the rigrht angle,
[SuG. Describe a circumference o
triangle as a diameter.]
Ex. 18. If two cirelea are tangent
tangeut touches them at A and B, re
right angle.
Ex. 19. If in the triangle ABC th
drawn, the augle ASD equals the angle AEB.
[Sua. Describe a semieireumference on A
righL
1 the hypoter
ixternally at !', and :
pectively, the angle
altitudes JIV and AE a
diameter.]
268. Def. a maximum is tlie greatest of a class of
mawiiitudes satisfying eertalE given conditions, and a
minimum is the least (see Arts. 470, 471) . For instance, of
the chords in a given circle, which is the maximum 1
EXERCISES. CROUP 18
MAXDIA AND
Ex. 1. 0! the chorda drawn through
determine which is the greatest, and all
Ex. 2. Find the shortest line, and
also the longestline, that can be drawn
from a given external point to the eir-
eumterenee of a circle.
[Sl'G. O beinp the center, prove
in A PCO, PA < PC ; by A PDO,
PB > PD.]
Ex. 3. Find the shortest line, and al
est line, that can be drawn to the cii
of a circle from a point within the circle.
[SuQ. O being the center, prove, by
A Ol'C, FB < PC. etc.]
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1-i'J liOOK II. PLANE l.!E
Ex. 4. If two cirples intersept, show
tU»l, of Hties drawn throufrli a piiict of ,
iJii-'ifei^tifiu and fenainnted by ihe> cir-
(tumfereiitt'S, that line is a loaximiim
which is pai'allpi to tlm
liueof <^e»t6rs.
[Sun. Prove US < 00' .: CPD < AI'B .]
Ex. 5. Given AB 1 Oli in eirele O ; prove
ZO.rBtbe ffiasimuraof all i having their vertieea
im the oivcumferecee and their sidespasHing througli
O and i! roepeetively.
a eirelo on OA as a diameter.]
EXERCISES. GROUP rs
DE MO XST RATIO XS BY IXDIRECT MKTTIODS
Prove the following by an indirect method (see Art. 195) ;
Ex. 1. A segment of a circle which contains a right angle ia a
semicircle.
Ex. 2. If a rectangle be inscribed in a eirelo, its diagonals are
Ex. 3. Prove the second part of Prop. VI by an indirect method,
Ex. 4. Prove Prop. X by an indirect method,
Ex. 5. A straight line connecting the midpoint of a chord and the
midpoint of the arc subtended by \\\e chord is perpendicular to the
chord (use the method of eoineidenee).
Ex. 6. A line joining the midpoints of two parallel chorda passes
through the center.
[Sl'g. Draw a J- to each chord from the center and show that
Ex. 7. If the opposite angles of a quadrilateral are supplementary,
a circle oan be circumscribed about the quadrilateral,
[SUQ. Pass a circumference through three vertices of the quadri-
lateral; it it does not pass through the remaining vertex, etc.]
Ex. 8. Prove the converse of Prop. XII,
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EXERCISES . CONSTAKTS AND LOCI
un
269. A geometrical constant is n, gfom(;trteal inagiii-
tude wliifih varies in some respect, as in position, but
remains constant in size. Thus the angles inscribed in a
given semicircle vary in position but are alt of the same
size; viz., a right angle (see Art. 261).
EXERCISES. CROUP 30
XATIOX OF CON'STAXTf^
AXD LOCI
Pis any point
Ex. i. All and AC are tangents to a eirck
cirouralerenee outaids tlie triangle ABC. As
sura of tiie Z.-iand / JiPC is conetant.
Ex. 2. In Es, 8, p. 12n, if AR and BQ be produced to meet at T.
Bhow that tlie venmeter ot tho triangle Tffy equals tlie sum of TJ
and TB, ami heiiee that the perimeter of triangle TBQ is constant, no
matter how /' may vary in position betmeiin -J and IS.
Es. 3, Shoiv o
the
■e that, i£ O
i the c
, Z.ROQU
Ez. 4. Two circles intersect in the points
A aod B. Prom any point P on one ciraum
fereuee lines FAC and PBD are drawn, ter-
minated by the other eireumference. Show
that the chord CD is constant.
[St-G. Draw BC and prove ICBD a,
Ex. 6. Given OA J- OB, and Cn :
(liven lungth moving so that D is alwa;
and C in OH and 7' the niidjiohit c
that OP is conf^faiit in IciikMi {-.c-e V
The dftermmatioH of ha
fucilitaied hij shoiviiig that
magnitude is a conslunt.
Es. 6. Find the locus ot a poi
dlataiiCL* a troui a giveQ circumtei'i
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144 isOOK II . I'LANE r.EOMETUY
Ex. 7, Fiml Tho locus of Iho mUipoiiils ot the ni
Ex. 8. Find the locus of tho mi.lnoiiits of :in chordsi of n ftiven
leugtli di'awiL in a given eirole.
Ex. 9. Find tlie )oi>us of tiie vertices of nU rij^lit lriaJii;li'K liaviag
a giTen liypotouuse.
Ex. 10. Find the loeus of the raidpointa of all tht ohoi'ds drawn
[Sl'G. Draw a line from tho center to the given point and perpen-
diculars from tbe teirtcr iipou fhn
chords.]
Ex. 12. QP
length and
always in a
line. Find the
EXERCISES. CnoyP 31
THEOREMS PRO^TA) BY VAKIOTJS ^IKTHODS
Ex. 1. The lino which liiseeta the angle formed by a tang
a ehord biseefs the intercepted are also.
Ex, 2. An inscribed trapezoid is isosceles.
Ex. 3. Given TA and TI! tangents, are J7.'= T^
m", an Sll = s:,°, and are 1)0=150°; find all the
angles of the figure.
Ex. 4. If two tangents to a circle are parallel,
the line joining their points of contact ia a diameter,
[SuG. Draw radii to the points of contact and
use an indirect method of proof,]
Ex, 5. A rectangle cireiimaerihed about a circle ii
[Sua. Use the preceding theorem.]
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MIKCF.LLANEOUS EXERCISES
Bx. 7. Find the angle formed by thi
Bide o( an macrlbed square and the taugeii
through the vertex of the square.
Ex. 8. Prom an external point a seeant is drawn through the
Center of a circle, and also two other secants making equal angles
frith tbe first secant. Show that the seeanta last drawn are equal,
Ex. 9. AliC ia a triangle and on the side AB the point P ii
taken and on liC the point Q, so that angle Bl'Q equals angle C.
Show that a circle may ba circumscvibed about the quadrilateral
Ex. 10. Two oiiClBB are tangent to each other externally, and a
line is drawn thtough the point of contact terminated by the circum-
ferences. Show that the radii from the extremities of this line are
PMallel.
Ex. 11. If a
Wangle as di^u
triangle.
Ex. 12. The chord of an are is parallel to the tangent at the
midpoint of the are.
Ex. 13. I£ a triangle be inscribed in a cirele, the sum n£ the
angles inscribed ia the segments exterior to the trianijle is four right
auBles.
Ex, 14. Find the corresponding theorem for an inscribed
"juadrilateral.
Ex. 15. Find the loi^us of the centers of all circles passing through
two given points.
Ex. 16. If two unequal chords intersect in a circle, the greater
chord makes the less angle with the diameter through the point of
intersection of the chorils.
e of this theorem. Is the
Ex. 17. The .51
tbe hypotenuse am
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UG
BOOK TI. PLANE GEOJtETKV
touch the m-
iiflo FQR and
Ex. 18. The aidos AIS, BG and AC of u tr
Bcribed circle at the poiuts I', Q and U. Show
one-balf angle A are complementary.
[Suo. Draw radii from the center to P and IE. Then I PQR
IFOA, etc.]
El. 19. Fi-om the point in whi
angle meets the circumforeiieo, a ch
of the angle. Show that this chord equals
the other aide of the angle.
Ex. 20. Given AB a diameter, AP=
the radius, AD and PC iBngents; prove
A CED equilateral.
[St'G, Draw OC and CA; then in rt.
A OOP, C^ = radma, ZP^GO", etc.]
Ex. 21. Perpendiculars are drawn
from the extremities of a diameter upon
a tangent. Show that the points in which ihu perpendi
the tangent are equidistant from the cente:.
CONSTEUCTION PROBLEMS
270. Postulates. As statod in Art. 49, a postulate In
geometry is a coustrmitiou of a geometric figure admitted
as possible.
The postulates used in geometry, aru as follows (see
Art. 50) ;
1. Through any two points a straight line may he drawn.
2. A straight line may J>e extended indefinitely, or it may
be limited at any given point.
3. A circMmference may he described about any given
point as a center and with any given radius.
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CONSTRUCTION PROBLEMS 147
271. Tlie meaning of the postulates is that only two
drawing instruments are to he used in making geometri-
cal constructions; viz., the straight-edge niler and the
compasses.
With these two simplest drawing instruments it is
ble to be able to construct as many geometrical
272. Form of solution of a problem. The stiitoment of
a problem and of its solution consists of certain distinct
parts which it is important to keep iu miiid. These are
1. The general enunciation.
2. The particular enunciation.
(1) Given, etc.
(2) To construct, etc. (or some other construction
phrase, as "to draw," "to bisect," etc.).
3. The construction.
4. The assertion.
5. The proof of the assertion.
6. The conclusion (indicated by Q. v.. p., quod erat
faeiendum, "which was to be done").
7. The discussion of special or limiting cases, if such
cases oceiir in the given problem,
la the figures drawn in eonnection with probleins, tbe (lircii U)ies
»fe drawn as hearij lines, the lines required fts light liiu-s, and the
oasiliary lines ea dotted lines.
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.ANE r.EO.MF/J'RY
PROPOsri'io>; XXIII. Problem
273, From a given point wilhout a give)i line to draw
a perpendicular to the line.
Given the line AB and the point 7* outsiilo AB.
To construct a pRi-petidicular from thi^ point P to the
line AB.
Construction. With P as a center and with any oonve-
iiient i-iidins, describe an arc intersecting A B in two points aa
at C and D. Rost, 3,
From Cand J> as centers and with convenient equal
radii, greater than J CT>, describe arcs intersecting at Q.
Poet. 3.
[Assertion]. Then PE- is the 1. required.
Proof. P is equidistant from the points C and T>. Constr.
Also Q is equidistant from the points C and B. Constr,
.-. PR 1 CD. Art. 113.
(two joints, eadli eqiiklistant from the exlremifies of a thie, determine the
X Mseetor of the line) .
Q. E. F.
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CONSTRUCTION PROBLEMS
Proposition XXIV. Problem
given line to erect a per-
274. At a given point in
pendicular to that line.
Kg. 1 Fig. ,3 .
Given the point P in the line A B
To construct a perpendicular to the line A B at the point P.
Method I. Construction. From V (Fig. 1) as a center
with a convenient radius, describe an arc cutting off the
equal segments PC and PB on the line AB. Post. 3.
From C ond I> as centers and with equal radii, greater
than PI), describe ares intersecting at li. I'oai. ?..
Draw PR. l-oBt.l.
[Assertion], Then PR is the _L required.
Proof. Let the pupil supply the proof.
Method II. Construction. Take any point (Fig. 2)
without the line AB, and with OP as a radius, describe a
circuniferenee intersecting the line AJi a.t C. Post. 3.
Draw CO, and produce CO to meet the cireumference
at R. Draw KP. Posta. I and 2.
[Assertion] , Then RP is the ± require<i .
Proof. Let the pupil supply the proof.
Discussion. When is Method II preferable 1^
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BOOK II. PI,AXR GF.OWK'r:
Proposition XXV. I'koblem
275. To biseci a given liHfi.
Given the line AB.
To bisect line AB.
Construction. With A and B as centers and with equal
radii, greater than i AB, describe ares intersecting in
€ and t). Post. 3.
Draw the line CD intersecting AB in F. Post. I.
Then AB is bisected at the point F,
Proof. Let t]ie pupil supply the proof.
PkOPOSITION XXVI. PltOBLEJl
276. 2'o hisfct a given arc.
Given tlie arc AB.
To bisect arc AB.
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CONSTRUCTION PROBLEMS 151
Construction. With A and B as centers, and with con-
venient equal radii, describe arcs intersecting at G and Z>.
Post. 3.
Draw the line CT> intersecting the arc AB at F. Post. 1.
Then are AB is bisected at the point F.
Proof. Draw the chord AB.
Then CD X chord AB at its middle point. (Why?)
.-. CD bisects the chord AB, Art. 223.
{the X hiseulor of a chord jiimscs Oirovgh the <:':nUr and Inserts tlie arcs
siihteiKlal by the chord) .
Q. E. F.
Proposition XXVII. Probleji
277. To Used <t oitrii aii'jh.
Given angle A OB.
To bisect angle A OB.
Construction. With as a center and with aiij' conve-
nient radiu.H, describe an arc intereeeting OA at C and
OB at J). Post. 3.
With C and I> as centers and with convenient equal
radii, describe arcs intersecting at F. Post. 3.
Draw OF. Pest. 1,
Then lAOB is bisected by line OF.
Proof. Let the pnpil suppjv the iit'ooE.
)j.a.jf,
fii- Construct ail I of iO°,
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15a i^ook ii. i'laxe geometky
Proposition XSVIII. Problejh
278. At a ginen point in a glvrn .^Imi'jhl Vn-' fo c
struct an mujlr equal to a given amjlc.
Given thp Z A, iiod the point P iu the line liC.
To construct at the point P an angle equal to A A, and
having PCfoi- one of its sides.
Conatnictiou, With .4, as a center ami with any eonve-
nient radius, describe an arc meeting the sides o£ /.A
at D and B. Post. 3.
Draw the chord T^E. Post. i.
With P as a center and with a radius equal to AT>.
describe an arc cutting FC at F. Po&t. .t.
From P as a center and with a radius equal to the chord
DP, describe an arc intersecting the arc HF at G. Post. 3.
Draw- FG. Post. i.
Then Z GFF is the angle reqnired.
Proof. Let the pupil draw the chord FG and complete
the proof.
Q. E. F.
Ex. 1. On B given line oonBtmet a square.
El, 2. Construct an angle of 30°.
Kx, 3. H«nce eonBtruet an. angle of 15".
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CONSTRUCTION PROBLEMS iM
Pboposition XXIX. Peoblem
279. TliTough a given point without a given straight
line to draw a line parallel to a given line.
Given any point P without the line A li.
To construct a line through P II -4 7i.
Construction, Through P draw any convenient line CT)
meeting AB in C. Poat. i.
At P in the line CD construct / JJPF equal to Z PCS.
Art. 278.
Then EF is the line required.
Proof. Let the pupil snppiy Ihe proof.
Ex. 1. Tlirough n given point dvaw a lino paraliel to a given line
by conBti'utiting a parallelogram of which the given point is one
Ea. 2. On a p;iven line as a diameter, oonKtniot a eirele.
Ex. 3. Oonatruet an are of 00" having a radius of 1 In.
Ei. 4, On the 9gnrB, p. 148, if the point Q be constructed below
the Una JB, will ihe perpendicular required be likely t« be more
accurately, or less accurately constructed ?
Ex. 5. From a given point on a piven eireumference, how many
equal chords can be drawn ?
Ex. 6. Throuf;h a p;iveQ point within a given cirenniferenee, how
many eqnal chords can bo drawn?
Ex. 7. Froru a given point esternal to a given oircle, how many
equal secants can be drawn !
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r,ANK GEOMETllY
PiutrosiTiON XXX. Ph(
280. Tn divide, n tjircn s/migM lim
nHiiiber of cqind inni.-<.
Given tiie line A B.
To divide AB into a required unuiber (as Uiree) eqoal
parts.
Construction. From ^1 draw the line AT iiutkiiig a con-
venient angle with AB. Post. i.
Take AC any line of convenient length and apply it to
AP a number of times equal to the number of parts into
■which AB is to he divided. Post. 2.
From E, the end of the measure when last applied to
AP, draw EB. Post. 1.
Through the other points of division on AP, viz., C
and Ji, draw lines ll EB and meeting AB at F and G.
Art, 270.
Then AB is divided into the required number of parts
at F and 6.
Proof. AG = CD = DE. Constr.
.-. AF=FG ^ GB, Art. 176.
(if three or more parallels intercept cq^ial parts on one transucrsal, Ihey
intercept e<pial partu on evenj traimversal).
Ex. 1 . CoDstruet a right triangle whose legs are 1 in. and li in
'Ef. %. Congtriiet a J ec tangle whose base 13 1 in. an<3 altitude 1 i:
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COKSTEDCTION PROBLEMS IDa
Proposition XXXI. Problem
281. To construct a triangle, given two sides and the
included angle.
Given m and n two sides of a ti'iuugle aiul P the augle
ineiiided by them.
To construct the triangle.
Construction. At the point A in the line Afl coustvuet
I A equal to the given /.P. Art. 2TS.
On the line AJi lay off AO equal to wi. Post, J.
On the line AF lay olF .) I) equal to *i. Post, ••.
Draw J)C. Post, 1.
The A ^I>C' is the triangle required.
1). E. F.
Ex, I. Constniat a triangle in ivhk'h two of the sides are 1 iu. snil
Itin,, nod the included angle is 13:)".
Ex. 2. Construct an isoaeoles tiiun^lu, in wiiiHli Ihe liase shall bii
H in, and the altitude 2 in,
Ex, 3. Constniet llie complement of a given acute angle,
Ex. 4i, Constniet the supplement of a given angle.
Ex. 5. How is tiio figure on p, 154 i^onstrueteJ wilh Ihe fewest
adjustments of tlie eompasses ?
Ex.. 6. DniK a. line (segment) iiud mark olY lliree-iifllis of it
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156 BOOK n'-. PLANE GKOMETRV
Pboposition XSXII. Problem
282. To consirud a triangle, givni tm angles and ihj
included side.
Given the A P and Q and the inr^luded side m.
To construct the triangle.
Consttuction. Take any line AI) anil on it mark AB
equal to m. Post. 2.
At A construct an angle equal to / P. Art, 278.
At B construct an angle equal to Z Q. Avt, 278,
Produce the sides of the A A and B to meet at €.
Poflt. 3.
Then A ACB is the triangle required,
Q, E, F.
Discussion. Is it possible to construct the triangle if the
sum of the given angles is two right angles ? Why ? Is it
possible if this sum is greater than two right angles ? Why?
n wliiai two of the angles are 30° anii
9 II iii,
Ex. 2. Construct the complement of half a given angle.
Ex.3, Coiistruot anaDgleo£120°; of I50°i o£ 135°.
Ex.4. Trisect a given risht angle.
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construction pkoblesis
Proposition XXXHI. Problem
;S. To construct a triangle, given the three sides.
Given m, n tmdp, the throe sides of ;i triangle.
To cottstruct the triiingle.
Construction. Take the line AB equal to m. Post. 2.
With A as a center and with a radius equal to n describe
an are, and with B as a center and with a radius equal to
p describe another arc. Post. 3.
Let the two arcs intersect at the point C.
Draw CA and GB. Post, \.
Then A ABC is the triangle required.
Discu^ion. Is it possible to construct the triangle if one
of the sides is greater than the sum of the other two sides?
Why?
What bind of a figure is obtained if one aide equals the
Bum of the other two sides 1
Ex. 1. Cotisti'uet ac aiiijle of 22-i°.
Ex. 2. Divide a given cireumlereiiee into four qundranta.
Ex. 3. Construct the figure on page 82, using tin
the tbcee altitudes aa & teat of the accuracy ot tbe woi
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158 BOOK II. PLANE GKOMETltV
PiiOPOSiTioN XXXIV. Problicm
284. To construiit <i irkuujh:, given two sidr-f and an
angle opposite one of them.
Given »i and n two sides of a A and Z 1' opposite n.
To conatmct the triangle.
Construction. Several cases occur, according to the rela-
tive size of the given sides and the size of the given angle.
Case T, Wlim n > m {<uid I V U acute) .
At the point A construct Z EA1> equal to Z P. Art. 27s.
On AE take AB equal to ?h . Post. 2.
With £ as a center and with a radius equal to h, describe
an arc intersecting AD at C and C.
Draw BO and BC.
Post. 3.
Post, 1.
Two A, ABC and ABC, arc obtained, containing the
sides m and n; but only one of them, A ABO, contains
the ZP.
.'. A ABO is the triangle required.
Case 11. When n = m (and /.Pis acute).
Make the same construction as in the preceding case.
The arc drawn intersects the line AD in the points A
and 0.
Hence the isosceles A ABC is the triangle required,
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CONSTKUCTION PE0BLEM8
Case III. When n < m {and IP is acute).
Make the eonstruetiou in the same way as in Case I.
Two ;&, ABC and ABC, are obtained, each of whieli
contains the sides m and n and an angle equal to / P oppo-
site the side n .
:. A ABC and ABC are the triangles required.
DiscussioB. In Case I, if ZP is a right Z , let the pupil
eoustruet the figure and show that there are two ^ answer-
ing the given conditions. If ZP is an obtuse angle, let
him construct the figure and show that there is hut one
answer.
In Case II, if Z P is right, or obtuse, what results are
obtained ?
In Case III, if ZP is acute and n = the ± from B to
AT), how niaoy answei'S are there? also, if w < this ±,
how many ?
If / P is right, or obtuse, what result is obtained ?
Ex. 1. Constniet a triangle in mhieh two of the sides are 1 iu, aud
ii in., and the angle opposite ibe latter side is 45".
Ek. 2. Construet a triangle in which two of tiie sides are IJ in.
and 1 in., and the angle opposite the latter side la 45".
Ex.3. Construct a triangle in which two of the sides are U in.
snd i in., and the angle opposite the Utter side is 30°.
Ex. 4. Construct the figure of page SO, using the coneurreuee of
the throe blseutois as a test of (he atjcuraey of thu woik.
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HOUK II. rLANE OEOME'J'ltY
pKOfOSlTlOX XXXV, PifOllLKM
285. To construct a -ihi rail dwj ram, i/Uru /iro sides and
the inclmkd uiitjle.
Given m and ii two sidos, arid P Uh' iiidiuli-d /. of a ITJ .
To construct the parallelogram.
Construction. Take line AB equal to m. Post. 2,
At the point A eoiistrnet /.OAB equal to ZP. ah, ?7m.
On the aide AG lay off AD equal to ".
From D as a center and with a radius equal to ))i, Luui
from B as a eenter with a radius equal to n, describe arcs
intersecting in E. Post. 3.
Draw ED and EB. Post. 1.
Thun ABED is the parallelogram required.
Proof. Let the pupil supply the proof.
Proposition XXXVI. Problem
286. To circumscribe a circle about a given triangle.
Given the A AliC.
To construct a circumscribed O about ABC
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COXSTEUCTION PliOIiLEMS 161
Construction. Ei'oct Js DFaud EG at tliu niidpoiiits of
the sides AV auil AB, respectively. Art. 274.
From 0, the poiiic of intersection of these A , witii a
radius equal to OA, describe the O ABC. Post, -i, Art. iST.
Then O ABC is the eii-cle required.
Proof. Let the pupil supply the proof.
Q. E. F.
PKorosiTiox XXXVII. Problem
287. To inscribe a cinie in a <jhcii Irianffle.
Given the A ABC.
To construct an Ui^^crihed circle in the triangle.
Construction. Draw the Hue AD bisecting ^EAC, and
Ci; bisecting- /lECA. Art. 277.
From 0, the iutersection of AT) and CE, draw the liiie
OP ± AC. Art, 273.
From as a center with a radius OP, describe a O .
P^-st. 3.
Then circle ia the circle required.
Proof. Let the pupil [supply tlic proof.
Ex. 2. 1
the tLree \
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162 book 11. plane geometry
Peopositiox XXSVIII. Problem
288. Throiujh a liiinii jicint on the circumfarence to draio
a tanijmt to a cirvk.
Given any pniot P on the (jircumference of the circle 0.
To construct a tangent to tlie cirele at the point /'.
Construction. Draw the radius OP.
At the point P construct a line AB X OP. Art. 374.
Then AB is tlie tangent required.
Proof. Let the pupil supply the proof.
289. An escribed circle is a circle
tiuigpiit to one side of a triangle and
to the other two sides produced.
Thus the cirele ia an escribed
circle of the A ABC. A center of an
escribed circle, as 0, is called an ex-
center Df tho triangle.
Es. 1. Draw a triaagle nnd all of its cscribtJ cirtlRs.
Ex. 2. Conatnict the figure on page 83, using tlie concurrence of
the three medians as a Itst of the accuracy of the work.
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COXSTKUCTION PSOELEMS lod
Proposition XXXTX. Phoblem
290. From a giw.n poini tvithout u circle to drain a iau'
gent to the circle.
Given F anj- point without the circk 0.
To construct a line through F taiigeot to the circle 0.
Construction. Draw the I'me PO. Post, :
Bisect the line FO at ^f. Art. 37i
From M as a center with Jff nri n
cumfurcQcc iuterscctiiig the giveu
and B.
;, tkseribe a cir-
iiftreiiee at A
Post. 3.
Draw FA and FB.
Then PJ ami PIS are tlie tangents required.
Proof. Z PAG is iLi>eribea in a semicircle.
■■. IPAO is a ri^rht angle.
.-. PA is tangent to tJie circle 0.
In like manner PB is tangent to the circle 0.
(Why!)
(Why?)
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164 BOOK II, I'LAXE r.Eo:in:TriV
Proposition' XL. I'lionLKM
291. Upon a given draUjM liiif to tlr^cribe a segmeni
irhich shall conlitin a gh-eii- iiifcrihed angle.
Given the ritraight line ^1 n and the Z P.
To construct on the line ,17>' a segment of a circle such
that any angle iuseribed hi tlie set^uieiit shall equal ^P.
Construction. At the point B in the line AB constract
ZABO equal to ZP. Art. 273.
Constract DE the ± bisectoi- of line AH. Art. 274.
From as a centev with OB as a radius, describe the
circle AMB. Post. 3.
Then AMB is the required segment.
Proof, Let AQB be any Z inscribed in the segment
AQB.
Then /.AQB is measured by h are AFB. Ai-t. 258.
But BG is tangent to the circle A2IB. (Why?)
.". I ABC is measured by \ arc AFB. fWhy?)
.-. /Q= lABG, or IP. (Why?)
.*. any Z inscribed in segment A2IB ~ ZP.
Q. E. F. .
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constkuctiox problems
Pkopositios XLI. Theorem.
292. To find the common unit of measure of ti
surable straight lines, ami hence the ratio of the lines.
Given two lines AB luui C7>, of wliieli CT> is tlie shorter.
To construct ii common unit of muaaiiro of J U luid CD,
and iieiJCR obtain tlie rutb oJ: AB rmd CD.
Construction. Apply CT> to AB as many times as
possible; Post. 2,
Say twice, with a remainder KB.
Then apply KI^ to CD as many times as possible ; Poet, 2.
Say three times, with a remainder LD.
Apply Lll to KB as many times as possible; Post. 2.
Say three times, with a remainder I'B.
Apply PB to XD as many times as possible; Post. 3.
Say it is contaiQed in LD exactly twice.
Then FU is the common unit of meiiiuiu of -I /.' :i\v\ I 'H.
Proof. LD=2PB.
KD=^KP + PB^3LD + PB=1PB. Axs. G, 8.
GD= CL + L7)^3KB + i/)=23P7i.
^ B = .4 71 + /r /J = 2 CD + KB = r,3 I'B .
A n _ r.r{ 7*/; „ :y.\
Hence -^ - ijj-^. = ^3"
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SXEfiCISES IN CONSTRUCTION PROBLEMS
293. Analysis of problems. The niothcxl of analysis
(see Art. 196) is of especial value in tlii; solution of con-
struction probleids. In general, to investigate the soiution
of a problem by tliis method;
Draw a fi<juri; in irJilch the vqi'irfil coiiKli-wtion is
ansumed as made ;
Draw auxiliary lines, ifnecflssriry ;
Observe the rvlaiioHS hdwceii the purtg of this fujiire, in
order to discover a known relaiion on ichiclt the required
construction depends;
Having discovered the required relation, construct another
figure hy the direct «se of this relation.
Es. Through a, given point within a oii'ole draw
a chord which shall be bisected by the given point.
Analysis. Let J be the given point witiiinthi?
given circle 0, and let FQ be a chord bisected at
the point A. A bisected chord suggests a line OA
joining tbe point of bisection with the center 0,
and that (Art. 224) OA ± PQ.
Synthesis, or Dihrct Koluteon. Taking miotlipv IlKure contain-
ing the data of the problem, connect the point A with the center of
theeircle by the line OA.
Through^ dniW a line 1 OA [Art,27+i, nnd meeting the circumfer-
ence at tlie points I' and Q. I'Q is tho thurd rL-tinited.
EXERCISES. CROUP SS
CONSTRUCTION OF STRAIC.nT LINES.
Ex. 1. Draw a line parallel to a given line, and tangent to a given
[Sua. Suppose the req^uired line drawn; then the radius to the
point of tangeney, if produced, is X given line, etc.]
Ex. 2. llraw a Hub porpendicular to a given line, and tangent to
& given circle.
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EXERCISES. PEOBLEIIS 167
Ei, 3. Prom two points In the eireumturenee of a eirele, draw
two equal and parallel chords.
Ei. 4. Through n giren point draw a line which ehall make a
given angle with a given line.
Ei. 5. Through a giveu point draw n line which shall make equal
angles with the sides of a given angle.
[Sua. The bisector of the s-Lven I will be ± the required Una, etc.]
Ejt. 6. Through a given point between two given pai'allel lines,
draw a line of given length with its extremities in the
two parallel lines. /'^^'""\
Ex. 7, Through a given point J within a eirele, IA/ "•, \
draw a chord equal to a given line. V \ \ \
Ex. 8. From a given point in the oircumferenco V j_.^ <- 'J
ot aoircle, draw a chord at a given distance from ^v.-^^!,-^
the center.
Ex. 9. Througli a given point on the circumference of a circle,
draw a chord which shall be bisected by another given chord.
[SuG. Draw the radius to the given point and on it as a diameter
describe a circle, etc. When is the solution impossible ?]
294. Construction of points and of loci. In construct-
ing a point to meet certain glvec conditions, it is often
helpful to c<nisirw:i the locus of a point iinsu-erhig one of the
given conditions and observe in ivhat point or points it meets
a given line, or meets anotJier locus ansirei-ing another
given condition.
EXERCISES. CROUP 13
COXSTBUCTIOX OP POIVl':: AXD LOCI
Ex.1. Find a point Pin a given line Jii ^ p
equidistant from two given points C and 11 '\^
[Suo. Construct the locus of all points ^
equidistant from C and D and observe where A~=^ \;__„^-_=^
it intersects the given line AB.] '
Ex. 2. Find a point 1' in a given circumfuicuce, which is equi-
distant from two givi'n iioiritB, (.' and D.
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1 givfu line, which i>
Ex. 4. Find a point
d, from a giren point.
Ex. 5. Find n pnint whioli is at tl given ilistanee, a, from a givfn
point A, riiiii at another distance, h, from anothiT givon i)oiiit B.
Discuss llio limitations of tliis problem.
Ex. 6. Find a point equidititant froia two givnn pointi^, mid at a
given distance from a given straight line.
[Srr.. Draw the locus of all points equidistant fi'ora the two given
points, and also the locus of all poluta at the given distance from tho
given straight line, etc.]
Ex. 7. Find a point equidistant from two given points, and at a
given distance from another given point.
Ex. 8. Find a point equidistant from two jriven points, and also
from two given intersecting lines.
On the other hand, the determiaatioa of certain loci is
equivalent to the conslrnciloii of all points which snti.tfy one
or more given conditions.
Ex. 9. Find the loeua of the center of a aivole, which touches a
given line at a given point.
[Sro. Construct a number of circles touching the given lino at
the given point and observe the relation of their centers.]
Ex. 10. Find the locus of the center of a ciraumterenoe with t
given radius, r, which passes through a given fixed point.
Ex, II. Find the locus of the center of a circle, touching two giver
intersecting lines,
Ex.12, Find tiie iocusof the c
parallel linos,
Ex. 13, Find the locus of the (
which toucliBS a given straight line
Ex. 14, Find the locus of tiic t
which touchcii a given circle,
ir of a circle of given radin:
■c of a eirale of given radiu
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EXERCISER. EROBLEJIS
EXERCISES. CROUP 24
COXSTRIXTIOX OF KECTII.INEAK riGL^RE
Ex. 1. An eijuilateral triangle, given the pecimeter.
Ex. 2. An eqiiilaloral triangle, given the Rltituda.
Ex. 3. An isosceles triangle, given tho bnse ami .iltiti
Ex. 4. All isoseeles triangle, given the ba.=o mul nri i
?cles
ri:mj;'l.,
Ex. 6. A right tviangle, given n. leg and iho acute angle adjacent.
Ek. 7. A right triangle, given li Ic},' and the acute angle opposite.
Ei. 8. A right triangle, given tlio hj'potenuse and an acute angle.
Ex. 9. A right triangle, given the hypotenuse and a leg.
given the altitude and the sides including the
Ex. 10, A
vertical angle.
[SUQ. Through tlie foot of the altitude d
of indefinite length, etc.]
p 1 altitude and
Ex, 11,
igle, g
iiJes and the
ilUtudf?
ipon
e of
Ex. 12. A square, given the diagonal.
Ex. 13, .4. rliomljHs, given the two diagonals.
Ex. 14. A I'homljus, given one angle and one diagonal.
Ex. 15. A parallelogram, given two adjaoent sides and an altitude.
Ex. 16. A parallL'lograr
.1, giv.
;n a side, the aUltude upon
that side
and an angle.
Ex, 17. A paralUdit-rt
im, yi
ven the diagonals and
an angle
included by tlK^m,
Ex, 18, A quadrilalerul
, b'ive
I, th« sides and one angle.
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170
BOOK n. PLANE GEOMETRY
295. Use of auxiliary lines ia constructing rectilinear
figures. In constructing polygons, auxiliary lines are fre-
(iuentlj' of service. Tinis it is often of especial value to
c<mstrtici, first, eitlier the inscribed or the eircumseribed
circle, and afterward tJie required triangle or quadrilateral.
Construction. Drawaeirolo with radius
equal to r. Draw a tangent at any point A.
On this tangent mark oft All and AC eiicli
equal i b. From B and draw tangents iJl''
and CF to tlia circle. TUen BCF is t!io
required trianglo.
EXERCISES. CROU^ SB
CONSTHUCTIOXS. AUXILIARY LINES
Coiistruet
Ex. 1. An isosceles triangle, givon the base and the radius of the
cireurasoribed eiraie.
Ex. 3. A right triangle, given (he radius of the iiircumscribed
eirele and an acute angle.
Ex. 4. A right triangle, given the radius of tlje Inscribed oirele
and an acute angle.
[SUQ. Draw the inscribed circle and at its center constrnet au
anele equal to the supplement of the given angle.}
[. 6. A triangle, given the base, flio altitude and tho vertex
-ingle.
a construct n segment which sliall contain
[Boo. On the given
the given vertes angle. See Art. 291. J
Ejc. 6- A triangle, given the baso, the mL'dian to the base, and the
vertex angle.
Ex. 7. A triangle, given one
of the inscribed circle. tSue Kx.
angle,
nd the radius
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EXERCISES. PROELEMH 17X
II adjacent angle, and the
The use of mtxiliary straight lines may be illustrated as
follows :
Ex, 9. Constmct a triangl6, ei^'^i the perimeter and two angles.
Analysis, Suppose the required triangle JBC already eonstriioted
and let AAISC and JCifbe the given
angles. Produce £C to D and E, ^
making DB = AB and CE =^ AC. -■-'/'^V-
Tban DE= given perimeter. Also --'"'/ \ '""""-
Similarly IACB=21E. Hence D B C £
CoKSTBUCTiON, Take D£tlie given perimeter; at 75 ctingtruot an
angle = X of 0"6 given angle! "t E eouatriict an angle = % ot the
other given angle. Produce the sides of these angles to meet e,t A.
Construct ADAB = ID and ICAE = IE. Then A ABC is tbo
required triangle, ete.
Construct
Ex. 10. An iaoscelea triangle, given the perimeter Kud the
altitude.
[Sue, Bisect the perimeter and eonstruet the altitude ± to it at
itsmidpoint-1
Ex. 11. An isosceles triangle, given the perimuter aud the vertoi
angle.
[SUQ, If the vertL'S Z is known, tiie base i may be obtained.]
Ex, 12. Arighttriangle,eiyonnnaciitH ^
angle and the sum of the legs. •' t"""--^
[SOG. Given AB the sum oC the legs, ,-■'' ^''~~--^.^^^^
eonstruet iiA=ii)°, etc.] _^ j) - ■- ^
Ex. 13, A riglit triangle, given an ueuto angle and the difference
ot the legs.
Ex. 14. A right triangle, given the hypotenuse and the sum of
the legs.
Ex. 15. A riiilit triangle, given au a«utu angle iind Ihs auoi o£ the
liyjotenuse and unu kjj.
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172 liOOK II. rL.VNR GEOJIT-.TRY
Ex. 16. A triangle, given an s.ng\-:, a siile nM thn sum of ti
Other tTTo sides.
Ex. 17. A ti'inngle, given an angle A, the sum o£ tlie sides j
and BC and tlie altitude upon AB.
296. Reduction of problems. Iti many caao.s a prolxhn
may be solved hy reducing the proilcm to a problem alreaii
soloed. (Tliis is a special kind of analysis.)
Kx. Construct a parallelogram, giycn the diagonals and one side.
Analysis: Suppose the £ZJ ABCF to bo Ihfl required /H? alread
constructed. Let AF be the given side, If the diagonals ni
given, halt each diagonal ia given (Art,
161). Hence in the A AOF the three j?
sides are given. Heneo the recinired / "^-.^ _,-■
problem reduces to the problem of con- / -''n'-
structing a triangle whose three sides •..--''''
.are given (Art. HS3), Henee
Construction. Let the pupil supply tho di
EXERCISES. CROUP 26
Rl'IDrCTION OF CONSTRUCTION PROBLEJIS
Construct
Ex. 1. A right triangle, given the aitituily upon tliy iiypotenusE
and ttie median upon the sniue.
Ex. 2. A rectangle, given the perimeter and a diagonal (see
Ex. 3. A rectangle, given the perimeter and an angle made by
the diagonals.
Ex. 4, A triangle, given the Ihreo angles and tlie railius of th»
circumscribed circle.
[SuG. The sides of the A are the chords of 11
segments of the O eontaiuing the given i .]
Ex. 5, A triangle, given two aides and the median / /
to the third side. j/'
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EXEKCISES. PROBLEMS li6
Ex. 6. Atrianftle, Riven the thi'ee medians.
[Suo. Beauee this to the pveeeding Ex. See Art. 187.1
Ex, 7. An isosceles trapezoid, giveu the buses and ati angle.
Bs. 8. An. iaoaceies trapezoid, giveu the biiaes and a diagonal.
Es. 9. A trapezoid, given tlie four sides.
Ex. 10. A trapezoid, given the bases and tiio two diagonals.
[Src. Reduoo to Art. 2S3 by producing, tlie lower base a distance
equal to the upper base, etc.]
297. Construction, of circles. The consfcraetion of a
required circle is freciuently a good illustration of tlie
preceding method of reducing one eonsfcruction problem to
another. For the construction of a circle frequently re-
duces to tlie prohlem of fiitdhuj a poitit {the center of the
circle) which answers giren eonditioiis. (See Art. 294.)
Ex. Construct a circle which shall touch two given intersecting
lines and have ita center in another given lino.
This problem is equivalent to the problem of finding a point which
shall be in a given line and be equidistant from two other given
lines. (See Ex. 3, p. lOS.) In soma cases, however, the eonstrno-
tion of a requirud t^irolii iiiiift be made by an independent method.
EXERCISES. CROUP 17
COSSTEUCTIOX OP CIRCLES
Construct a circle with given
, radius, r,
Ex. 1. -Which passes thr.
line.
ough a given pfiint and tf
Kiches a given
Ex. 2. Which has its ce'
uler in a given line uud t,
luehes another
given line.
Ex. 3. Which passes thri
)ugh two given points.
CocHtruet ft cirelo
Ex. 4. Which touches tw
•0 givou parallel lines and
passes through
a given point.
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Ex. 5. "Wiiii.-li parses llinin^'li two gi^-Lii
on a given line.
Ex, 6. Whiuh toucho3 three given lines,
two of which are paraHcl.
Ei. 7. Which passos through a givon
point A and touches a giren lice liC at a
given point J>'.
[SOG, Draw An tmd at li coiiatriLfl ;i 1
to fiC]
Ex. 8. TV'hicli toiithes a givenlins and al:
touches s, given circle u
/ \ a given point ^.
..o
Ex. 9. Whieh touchefi .i given line AJl
t a given point A and touches a given
EXERCISES. CROUP 28
PROBLEMS SOLVED BY VARIOUS METHODS
Kx. 1. Through a given point draw a line which shall cut two
given intersecting lines so as to form an isosneles triangle.
Ex. 2. Construct an isosceles ttiimgle, given the altitude and
one leg.
Ex. 3. In a given circumference find a point equidistant from two
given intersecting lines.
Ex. 4. Draw a circle whicli shall toui^h two given inlerseoting
lines, one of thein at a given point.
Ex, 5. Draw a lino which shall lie terminateJ by the sides of a given
angle, shall equal a given line, and be parallel to another given line.
1 ndjac
mgle,
Ex. 7. Find a point in a, given circumference at a given distanea
from Si given point.
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MISCELLANEOUS EXERCISES. PROBLEMS 175
Ex. S. Construct a parallelogcam, given a siJo, an angle, and a
diagoniil
Ex. 9. Through a given point within an angle, draw a straight line
terminated t>y the sides of the angle and bisected by the given point,
[SuG. Draw a line from the verte-i: of the angle to the given point
and produce it its own length through the point.]
Ex. 10. Construct a triangle, piven the vertt'i angle and the
segments of fhe base made by the altitude.
[SuQ. Use Art. 291,]
n ciidiiis which shall touch a
Ex. 11. Construet an
isoscclf
ertcxand the base.
Ex. 12. Drawacircli
5 with si
irele at a given point.
Ex 13. Construct a
right tri;
Iti tilde upon the hypoto:
QUSO.
■, given the hypotenuse and the
Ex. 14. Construct a triangle, given the base and the altitudea
npon the other two sides.
[SuQ. Construct a semieircle on the given base as a diameter.]
Ex. 15. Find a point in one side of a triangle equidistant from
the other two sides.
nd the angles at
Ex, 17. Construct a rhombus, given an angle and a diiigonal.
Ex, 20. In a given circle draw a chord equal to
parallel to another given line.
[SuG. Find Ibe distance of the given chord fro:
couattiieting a right triangle of which the hypotouuse
given,]
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1(G 1iOOK II, PI.AXE GEOMETKY
Ex. 21, Consfrupt a ti'in!if;Io, giv'ii nil aiL.ijii;, thu bisei'tor of that
angle, and tbe aititude from iiuotlitir vertox.
Es. 22. Find the loeus oE the points (jE ciontact oE tangL'tils dfawa
from a given point 1o ii hltIl's of circlea kaviii!!; a given i^tuter.
[Sm, Uao ArU. il'JD mid 2til.]
Ei. 23. Given a line ,17; ami Uvo points. -G
C and D, on the same side of JJl. Find a
point P in ^J! such that IAPC= IJU'D.
[Suo. Draw a 1 from C to J7! ajid pro- j^ +^ ^
dueeit its own length, ecc]
Ex. 24. Given a line -17i and two points r and }> on the samo side
of AB; find a point P in J7( suth that CP + I'D sliall ho a minimum.
Ex. 25. Draw a
tangent to two given eireh
Ex. 26. Dra'
tangent to two given circles.
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Book III
PROPORTION. SIMILAR POLYGONS
THEORY OF PROPORTION
298. Ratio has been defined, and its use briefly indi-
cated in Arts. 245, 246.
299. A proportion is an eKpressioTi of the eriiiality of
two or more equal ratios. As,
This reads, "the ratio of d to J equals the ratio of c to
d," or, "a is to i> as c is to d."
800. The terms of a proportion are the four quantities
used in the proportion. In a proportion
the antecedents are tlie first and third terms;
the consequents are tlie second and fourth terms;
the extremes are the first and last terms;
the means are the second and tJiird terras.
A fourth proportioaal is the last term of a proportion
(provided the means are not equal).
Thus, in « ; fe = c : d, dis a. fourth proportional.
301. A continued proportion is a proportion in which
each consequent and the next antecedent are the same.
Thus, a : & = & :c = c : d = d : ^ is a continued proportion.
A mean proportional is the middle term in a continued
proportion containing but two ratios.
h (IT")
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1(8 ISOOK III. PLANE GEOMETliY
A third proportional is the last term In .i coiitiimod pro-
portion containing but two ratios.
Thus, m n : h^h : c, b i& a. mean pi'oportioual, and c is a
third proportional.
Phoi'OSitios I , Theorem
302. In any proporiion, the prndiict of the extremes is
equal to tlie pvodwl of tlie wniiis.
Given tlie proportion a : b = r : d.
To prove ad = be.
Proof. f^ = 2- "^i"-
Multiply each memher by hd.
ad — he. Ax. i.
Proposition II. Theorem
803. TJie mean proportional iHween two quantUie;
equal to ike square root of I heir product.
Given tlie jtroportion a :l) = h :
To prove b = y(
Proof. a:l>^h :
:. h" = ac,
(in any projfortion, the i>rodiict of th-
.-. b=v;^
Hyp.
Act. 303.
equals the product of the
Q. E. B.
Ex, 1. Find the fourtli proportional to 2, 3 and C ; also to 3, i, ^.
Ex. 2. Find tlie mean proportional between 3 and 6.
Ex. S, Find the third proportioual to 3 and 5.
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THEORY OF PROPORTION 179
Proposition III. Theoresi
304. If the product of two quantities is equal to the prod'
net of two other quantities, one pair may be made the ex-
tremes and the other pair the means of a proportion.
Given ad = hG.
To prove
a , b^c: (I.
Proof.
ml- Ik.
Divide eacli memlici
l.y M.
Then
b' d
Or
o :S-C ; <!.
306. Cor. 1. If the (uitecedeuts of a proportion are
equal, the consequents are equal.
Thus, if a:x = a\y, then x = y.
Let the pnpil supply the proof.
306. Cor. 2. If three terms of one proportion are
equal to the correspondinrj three terms of another proportion,
the fourth terms of the two proportions are equal.
Thus, if II : h = c : x, and a : li = c -. y, then x = y.
Let the pupil supply the proof.
Ex. 1. 'Write <ih = i>q as a proportion in as many different ways as
Ei 2. Write a:(j:+l) = (i na a proiwtiou ; ulao j;- = 15.
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Proposition IV. Tueokem
807. If four quantities are in proportion, they are in
proportion 6j^ alternation ; that is, the firnt term in to the
third as the second is to tlie fourth.
Given the proportion a •.b = c -. d.
To prove a . c=h -. d.
Proof. a : h^c : d. Hyp.
.-. (id = bc, Art. 302.
(in anu pronortimi, ike product of the extrnauis equals the iiroduct of
Writing a and d as the extremes, and c and h as the
means of a proportion,
a : C = l) : d. Art. 304.
(*/ the product iif tieo qnaiitities is eqiiiil io ih^ prodni-t of iiro nlher
Proposition T . Theorbm
308. If four quantities are i» proportion, they are in
proporHow 6j/ inversion ; that is, the second term is to (he
first «s the fourth is to the third.
Given the proportion a ■.h = c -, d.
To prove Ji : a^—d -. c.
Proof. a : h — C: d. Hyp,
.-. ad^bc. (Why?)
Writing 6 and c as the extremes, and a and d as the
means,
I : a^d : e. (Whylj
Q. E, ».
Ex, TranB£ormi:a = i):o bo that j: shall be the last term.
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TlfROliY OF PROPOKTION 181
Peoposition VL Theorem
309. If four quanfitips are in proportion, they are in
proportion by compositioa ; iktif ix, the. snm of the first two
terms is to the second term ws the sum of the last two terms
is to the last-term.
Given the proportion a : h — c -. d.
To prove a -{- h -. h^c-V d: d.
Proof. 7= T ^?P-
h a
Add I to oach meiuber of the equality. As. 2.
h a
^ b d
That is a + h -.h^c + d: d.
Let tlie pnpil show also that a -r h : n = c-{- d : c.
Froporitios- VIT. Theorem
310. If four quindiUcs are in proportion, they are in
proportion ?!(/ division ; that is, the difference of the first
two is to the second as the difference of the last two is to
the last.
Given the proportion a -. b = c : d.
To prove a — b:l> = c — d:d.
Proof. ^ =^- Hyp.
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18'J HOOK iTt. i=LA>rE nroMi
Subtratit 1 £min vM-h mi>,m)Mv of I he •
1.1 d
a — h '■ — >!
That is fl-?i : i = c — d : (L
Q. E. D.
Let the pui>il hIkiw also that a^b • a—c — d c.
PRorosrnoN VTII. Theorem
Sll. If Jovr quaniities are in proportion, they are in
proportion hy composition and division; that is, the sum of
the first two is to their difference as the sum of the last two
is to their difference.
Given the propoi'tioii a -. b = c : d.
To prove a-\- b -, a — !> = <■ -\- d -. r, — d.
Proof. *' ; /' = '■ : 'I. Hyp.
^'-^h_c^d_
By cooi|msifii!
d
Dividing the coiTespoiiding men
equalities,
a+b^ c+d
a — h e — d'
That is a-^ h : a — & = c + d -.
Ex. 1. WhHt do«e the proi-iortion 12:^ = 8:2 beeoi
tionT also ty divisiout
Kl. 2. Whatdoea 2.c — 5:5 = 3^: — 7:7 becomoby t
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THEOBY OF PROPORTION lod
Proposition IX. Theorem
S12. In a series of equal ratios, fhe svm of all (he ante-
cedents is to the sum of all the consequents as any one ante-
cedeni'is to its consequent.
Given a -.l^c : d = e if^g : h.
To prove a-\- c-\- e-}- g -. J> + d+f+ h = a -. h.
Proof. Denote each of the eqnul ratios by r.
Then l^r, :. a^br.
Similarly c — dr, e=fr, g — hr. Ax. i.
Hence a + c + c + q={b -\- <l -i- f + h) r. Ax. 2.
Dividing by b i- d +f+ h, ^qrf+y:^ = *" ^ ^ ' ^'^- ^■
That is a + c-<re + g -.h + d + f+h^^aa.
Proposition X. Theorem
313. The protiucts of the correspovllut/ /ci-Jiis of two or
more proportions arc in proportion .
Given a : I'^c : d, e : f^g -. h, ami j : A-=i : m.
To prove aej ; hfk^cgl : dhn.
Proof. 7= -,. 7= f , T= -■ Hyp.
h d f h k. i)i
Multiplying together the uorvetipoiiilijig terms of these
equalities,
Ai. 4.
Q. E. B.
aej _ cgl
t)fk dh7n
That is aej : bfk^cgl : dhm.
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54 EOOK III. 1>T,ANF. CJEOMETIiV
Proposition XL Theorem
314. Lilie 'powers or like roots of ilie terms of a proper-
on are in proportion.
Given the proportion » -. h^c -. d.
To prove o" : V ^ c" ■. i!", and a" : h" = v" -. d".
Proof, 7= V Byp,
Raiding both loembers ui the h"' power,
That is a" -.h" =c" :(!".
In like manner ci" -. ?i" = c" : d".
Proposition SII. Theorem
815. Equimultiples of two quantities have the same
ratio as the quantities themselves.
Given the two c|naiitities a and 6.
To prove ■ma -. viJi = a -. b.
Proof. 7- J- Went.
Multiply eaeh term of flie first fraelion by m.
ina_ a
■'■ ml>~ h
That is ma : mh = a -. h.
__^_________ Q. E. B.
Ex. 1. Transform ji : q = x : j in all possible ways by the use of the
properties of a proportion.
Ex. 2. TraQB£ovm 7:3 = 28:12 by eompoaitiou aiiJ diTiBion.
Ex.3. Also2a; + 5;2i-5-a^+l:j;' — 1.
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THEOBY OF PEOPORTION 18.)
816. Note. In the theory of proportion as just p^eaented, the
CjuaQtities used are assumed to be ooinmenaurablo, but the saiue
theoreniB may also be proved for proportiona wlioae termB ara iucom-
menaurable by use o£ the method of limita. For each incommenan-
rable ratio may be made the limit of a eorrespondiug eommenaurable
ratio; then, by ahowing that the variable commensurable ratios are
equal, it may be proved that the limiting incommensurable ratios
It is also to be noted, that, in the above theorems, the terms of a
ratio must be of ttie same kind of quantity; that is, both be linea, or
both besurfacus, etc. Hence, in orderthat aproportion be treated by
alternation, for instance, all four of the terms must be of tlio same
Ex. 1 . Find the fourth proportional to a, 2<i, 3 ic.
Ex. 2. Find the third proportional to rt -;- '< and a — h.
Ex. 3. Find the value of x in the proportion, 4 : 5 = a ; 15.
Bx. 4. Findft shortmethodof determining whether a given propor-
tion is true or cot. Use this method to determine whether the follow-
ing proportions are true: (1) 4 r 6 = 3 : 9. (2) 5a : 2o = 15 : 6.
Ks. 6. Construct exactly the figure of page 77 with the fewest pos"
sible adjuatmenta o£ the compasses.
Ex. 7. Draw an obtuse triangle and
uaing the eoncuiTeuce of the altitudes as a t
Ex. 8. Arrange nine points in a plane
greatest number o£ straight liues may pasa i:
pass through three points and only three.
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EOOK III. TLAXR flFllMETP.Y
PRorosiTiox Xlll. Tin:oBKM
317. A line panilhl l'< ime, -ihlc of a triangle and meet
ing the other two skUs, dlfides these sides proportionally.
Given the tviaiiKle ABC ami the Urn- DK \\ base liC and
intersectintr the sides AH and AC in ilie points 1) and J-],
respectively.
To prove BB -. AD^EG : AE.
Case I. When T>B mid AD (Fig. 1) are commensxirabje .
Proof. Talie iiny cornnion unit ..f niKisiiro of I)B and
AD, as AS', and let" it be .'Oiitainu.l iu /)/.' a cerlain lunnWv
of times, as n times, and in AP, m \\mv^.
DP,
Tbroiigli the points of division of DP. and AB drai^
lines II BC.
These lines wiil divide EC into i\ . and AE into m parts.
all equal.
EC_n
•'• AE in
DB^EC
AD AE
Art. 17G.
(Why ?)
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PROPORTIONAL LINKW 187
Case II. When DB and AD (Fig. 3} are incommenKurnlile.
Let the line AB be divided into any number of equal
parts and let one of those parts be applied to DB. It will
be contaiued in DB a certain onraber of times, with a line
PB, less than the unit of measure, as a remainder.
Draw PQ \\ BG.
Then DP and AD are commensurable. Constr.
AD AE '^^'
If now the unit of nieaKure be indefinitely diminished,
the line PB, which is less than the unit of measure, will be
indefinitely diminished.
Hence DP ^ DB, and EQ ^ EC us alimiL Art.'isi,
DP
- AD
EQ
) ^^„.. .. ........ ^
DP ^, . -,-, EQ
— ^thevanablej^a
But the variable
;. the limit ^-^,= tho limit f|- Avt. ^.w.
AD AE
d. E, D.
318. Coit. 1. By eoniposition. Ail. ".09.
DP + AD : AD^EC -{■ AE : A E.
Or AB : AD = AG i AE.
In like manner AB -. DB = AC -. EG,
or, in general language, if a line parallel to the base cut the
sides of a triangle, a side is to a segment of that side as the
other side is to ilie corresponding segment of the second side.
319. Cor. 2. Using Fig. 2,
m^AL I!^=A^ PTJ^DP
QO AQ' ^^^^ EQ AQ " QG E</
Henee. if two lines are cut by a number of parallels, the
corresponding segments are proportional.
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188 IKKlK III. PLAXE GROMETlii-
Proposition XIV. Theohi-.m (Convehse op Vmn'. XlII)
820. If a stffiinht line diviiUn iiro si'h-n of a triangle
proportionally, it is iMraUel to the third side.
Given the A ABQ and the linu DF intersecting AB and
AG BO thut AB : AB^AG : AF.
To prove DF]] BG.
Proof. Through 7> draw the line BK \\ BG and meeting
the side -iC in K.
Then AB : AD = AC : AK, Art. sis.
(if a line II bam i^v,t lite sides of a a, " aide is to a segment of ilial side aa
tii^ other side is to the corresponiling segment oj the second side).
But AB : AD-^^AG : AF. Hyp.
.*. AF^AK, Art, 306.
[if three terms of oni! ^roporthm are cijiinl to the enrrespouding three
terms of another prn/wtiiHi, Ihr.fiiiirlli terms of She tiro '
j,roi>orUv)is arcapial).
Ilenee the point K falls on F, and the line 1)K coin-
cides with the line DF. Art. 6C.
But the line DK 11 BC. Constr.
,-. BF !! BG,
(for DF coincides Kith DK, ivkicii is || BC).
Q- E. D.
Ex. 1. In the figure of Prop. XIII, if AD = 6, DB = 4, and ^£=9,
find EC.
Ex. 2. Also, if AD=12, Dii = S, aad AC^la, find AE aaii EC.
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EXERCISES. CROUP IB,
REVIEW ESERCISKS
Make a list o£ the propertii':
Ex. 1. One straight lin
be related to the line).
Ex. 2. Two straight lines that meet, or intei
With such angles as may be formed by the given, li
Ex. 3. Two or more para,llel lit
and angles formed).
Es. 4. Eight angles.
Ex.5. Complementary angles.
Ex. 6. Supplementary angles.
Ex, 7. A single triangle (in
triangle, aa altitudes, medians, et
Ex. 8. A right triangle.
Ex, 9. An isosceles triangle.
Ex. 10. Two triangles.
Ex, 11, Two right triangles,
Ex. 12, A rjuadrilatersl.
with RHph points as may
nnneotion with lines within the
Ex. 13. A trapezoid.
Ex. 14. An isosceles tvapeioid,
Ex. 15. A parallelogram.
Ex. 16. A rhombus.
Ex. 17. A polygon.
Ex. 18.
A single
related poi'
flts).
Ex, 19.
Arcs of
angles) .
Ex, 20.
Chords 1
Ex. 21.
Central a
Ex. 22.
Tangents
Ex. 23.
Seeants 1
Ex. 24,
Two ciR'
niferei
(ir
rotated lines ai
^-ith related lie
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190 HOOK in. PLANE OEOJIETRY
SIMILAH POLYGOWS
321. Bef. Similar polygous are polygons having their
homolosroiis Liiigliis et^iial and their homologous sides pro-
portiooal.
Thus, if tliR figures AB€7)E and A'B'C'B'E' are simikj,
the angles A, IS, 0, etc. , mnst equal the angles A', B', C, etc.,
respectively; also ^B : A'B'^BC : B'C'^ GJ) -. OB', etc.
Hence it is cOiiMtantly to be borne in iniud that simi-
larity in shape or form of rectilinear figures involves two
distinct properties:
1, The liojiiologoiis angles are equal.
2. The homologous sides are proportional.
It should also be clearly realized that one of these
properties may be true of two figures, and not the other.
Thus, in the rectangle A and the rhomboid B, the cor-
responding sides are proportional but the corrt
angles are not equal.
!i
Also, in the reetangle G and the square J\ the corres-
ponding angles are equal but the corresponding sides are
not proportional.
However, it will be found that, in the case of triangles, it
one of the two properties is true, the other must be true also,
322. Def. The ratio of similitude in two similar figures
is the ratio of any two homologous sides in those figures.
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SIMILAR POLYGOKS
Pboposition XV. Theohem
323. If two triangles are mutually equiangular, they
are similar.
Given t!ip & ABO and A'B'C witli lA^ lA', IB =
AB', and / 6' = Z t".
To prove tlie &^ ABC and A'B'C similar.
Proof. Tlic given & are niutuaEly equiangular. Hyp.
Hence it only remains to prove that their homologous
sides are proportional. Art. 321,
Place the A ^.'S'^upon the A ABO, so that I A' shall
coincide with its equal, the ZA, and B'C take the posi-
tion FH.
Then lAFR = IB. Hyp.
.-. FHWBC. (Why?)
.-. AB : AF^AC : AR. Art. 317.
Or AB : A'B' = AG : A'C. Ax. 8.
In like manner, by placing the A A'B'C upon the A
ABC so that the Z S' shall coincide with its equal, the ZB,
it may be proved that AB -. A'B'^BC : B'C.
Hence the A J,£Cand A'B'C are similar. Art. 331.
324. Cob. 1. If two triangles have (wo angles of &ne equal
to two angles of the other, the triangles are similar; also.
If two right triangles have an acute anijle of one equal to
an acute angle of the other, the triangles are similar.
325. Cor. 2. If two triangles are each similar to the
same triangle, they arc similar to each other.
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192 BOOK III. PLANE GEOMiyi'EY
Proposition XVI. Tiieohkji
326. Jf iieo triauffles liavc their }w>nolo<jous sides pro-
porlional, Ihey ure. similar.
Given tlie A ABC and A'B'C, in which AB : A'li'^
AC: A'C'^BC: B'C.
To prove the A ABC and A'B'C similar.
Proof. The given & have their homologous sides pro-
portional. Hyp.
Hence it only remains to prove that the A are mutually
equiangular. Art. 321.
1. On AB take AF equal to A'B', and on AC take AS
equalto A'C. Draw Fff.
Then AB -. AF^AC -. AH. Hyp.
.-. FH II BC, Art. 320.
(if a siraiglit line dickies two sides of a A proportionally, il is || the
third side).
:. ZAFU = IB, and IAHF= I C. (Why?)
,•, A AFH and ABC are similar, Art, 323.
{if tico &. are muttiatly equiangular, they are similar) ,
2. .■- AB : AF=BC : FR. Art. 321,
Or AB : A'B' = BG : FH. A:^. 8.
But AB : A'B' = BC : B'C. Hyp.
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SIMILAK POI-TGONS
Hence the A AFH and A'B'G' are ecinai.
.-. A A'B'C is similar to A ABO.
[for ils equal A AFS is similar to A ABC).
Proposition XVII. Theorem
327. If two irianyles have an angle of one equal to an
angle of the other, and the inciiidhiy sides proportional, the
trianglea are Kimilar.
Given the A AHVuud A'B'C, in wiiit^li Z.A^ /.A' and
AH : A'B' = Aa: A'V.
To prove tlie A ABC and A'B'C similar-
Proof. Place the A A'B'C upon t.lie A ABO so that
ZA' shall' coincide with its equal, the Z.A, and B'O' take
the position FH.
Then AB : AP^AG : All. Hyp,
Hence FII \\ BO. (Why?)
.-. Z AFII = /.B, D.n<i ZAEF= ZC. (Wiy!)
.'. A ABC and AFH are similar. Art. asrt.
Or A ABC and A'B'C are similar. Ax. 8.
(). E. D.
miliirA.i'ii'O',
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Proposition XVIIl
328. If iivo tnangles have tK.. ^
pp.ndiciilar, each to each, the triangles are similar.
Given tlip A A'B'C (Fig. 2) with it;
Theorkm
S'i'les parallel, or per
and tliM
■espondir
A A'H'ff (Pig. 3) with its sidos -L,
nf the A ABC.
To prove & ABC and A'B'C similaf.
Proof. The A A and A' are either equal or supple-
mentary, {foi-llie»i<lc3fm-miig tJ,emn:-e\\ or X). Arts. 130,13?.
Similarly, the A B and B', and C and C are either
eqnnl or supplementary.
Hence one of the three following statements must be
true eoneerning the angles of the A: either
1. The A contain three pairs of supplementary A and
lA^ lA' = 2vX. A. IB -^ IB' ^2 rt. A, AC + AC' =
2 rt. A ; or
2, The A contain two pairs of supplementary A , as
/.4=ZA', ZB + Zit' = 2rt. A, I C + AC = 2n. A : t,r
:i. The A contain three pairs of equal A and AA =
AA', IB=IB'. AC=AC, Art. 139.
(if two A of a A = two A of another A, the third A are tqnai).
The first two of these statements are impossible, for the
sum of the A of two A cannot exceed four rt. A . Art. 134.
Hence the third statement is true, and the A ABC and
A'li tj MM liiutuaily cqiiiungular, and therefore Bimihir,
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Proposition XIX. Theorem
329. If iiro pohjijmin arc xiMilci-, ihcij may he separated
into the Sfime number of iriaHglen, similar, each to each , and
similarly placed.
Given tlifi similar polygons ABODE and A'li'C'JyE',
dividml into triangles by tho diagoiiids AC, AD, and A'C,
A'jy drawn from the corresponding vertices A and A' .
To prove tliat & ABC, ACD, ADE are similar U) tlie
^ A'B'C, A'Cry, A'D'E', respectively.
Proof. 1. /.li ^IB'. Ait.:m.
Also AJi : A'B'^Tin B'C. Art. r.'Ji.
;. AABV and A'B'C are similar, Art. yy..
U/Uco ii l,i(fc<iii L of one = <m Z of the i'lhe,- >„;l the ii,<-l,idim;
2. Again IBGI) ^ IB'C'D', Art.s-i.
(JiwB(0io(70ns i of similar poliigojin) .
Also Z BCA = Z i"C".4.'. Art. 321 ,
{homolo!i,ms A of tlie Similar A AlSt: <uid A'H'C).
Snbtracting lACD = lA'C'D'. An, a.
But £0 : B'G'=CD : C'Jf, Art. 31!i.
{homologous sides of siiiiiltir pnJi/goiis are proportioTial).
And TiC: /i'C = ^lC: ^'G', Art. 321.
(iMjijr homologous siili^s of the similur & ABCmd A'B'C).
Hence jIC : A'C ^ CD -. CD'. Ax. i.
:. the ^ A6'/J and A'G'iy are similar. An. 327.
3. Ill like manner it oau be shown that the A .iDE
i^nd A'D'E' are similar. q. z. b,
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BOOK in. PLA-NK lir.i:
PROK)tiITIOX XX. THEDliEJl (CONVKKSE or PliU!'. XIX)
330. // two poli/gons are cotiiposi'd of the name, nmnhvi-
of triangles, similar, each to each, and similurly placed, the
polygons are similar.
Givea the two polygons ABCDE and A'B'OB'W, m
■wliieh the A ABC, ACJ>, AD-E are similar, respeetivelj", to
the A A'B'C, A'C'JV, A'lyS', and are smiilarly plaued.
To prove the polygons ABCJDE and A'B'C'D'E' similar.
Proof. /.B = ZB', Alt. 321.
{hfiiKj homologous S of similar &.).
Also ABCA ^IB'CA'.
And lACD^lA'C'D'.
Adding PC(iials. ZBOl) = I B'C'iy.
In like manner it may be shown that Z Cl>E~ Z (■'!>' ?J\
ZBAE= IB'A'E', etc.
Hence the polygons are mutually equiangular.
., AB BG ,, . ., ^ .
^^^'^ .IIP, " "STT^ (homologous sides of mi
{Why?)
(Why-,
(WLy.^
Art. 321,
Again -
In like n
BC
B'C'~
CB
CI)''
(Why?;
A'B' B'C
BG_^AC_ CD ^AG
^ B'C A'C" C'ly A'O
CD ^BE __^A E
CD' D'E' A'E''
Hence the homologous sides of the given polygons an
proportional.
.-. the polygons ABGDE and A'B'C'D'E' are similar.
Art. 321
9. X. B.
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I'ROl'OR'J'lOSAL LINES
331, Note, It is often eonvsment to write a saries of (
ratios, like those uspd ia the above proof, us follows:
AB^^BC _/'AC'\ _ £D.^('£0^_ DE _AR
A'B' B'C \A'C'J~ azf yA'D')~ D'E'~ A'E''
a ratio which ia used merely to show the equality of two other r
being inclosed in parenthesis.
Proposition- XXI. Thkokem
332. In any triangle-, the bisector of an angle divides
the opposite side into segments lekich are proportional to the
other two sides.
Given the A A H(\ with the line A l> bise^^ti[lg the Z BA C,
Jitid meeting BC at /',
To prove ftC ; !>H^A<! -. AH.
Proof. DriiW the line CF\\AD, and moetiiijc ^1 />' pro-
dnee,d at F.
Then 1)€ : DB^AF : AB. Art, 317.
But Zr^lp'. Art. 124.
And ls= £p. Art. i:!ii.
Also £p'— Lp. Hyp.
.-. Lr= Is. (Why?)
.-. A .40Fi8 isoseeles, a.m\AC=AP. (Why?)
Buhstituting .-1 ('for itsec|uiU, A-F, in the above proportion ,
DC : DB^AC: AB. An-, s.
Q. E. B.
Ex. If. ill tlie above figure, AB^U.AC- VI. nud /ir r M, find DC
and Di:.
[SvG. Let lJC = Jr. /V/f-U — 3, etc.]
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)S HOOK .III. I'LANi; (;eu.metky
333. A line divided internally Ua Vuiv divideil into two
irtK by ii Doint ty.ken betvveen its extremities.
Tims, tlie line Ali is divided intpriially at I
into tl)e segments PA and PB.
To divide a givi'n line interjially in a frU-Pii ratio (as '
334. A line divided externally is ii line wliose parts ai
considered to bo llie segments included between ;t ]ioiiit (j
the line produced and the extremities of the gi^■cu line.
Thus, the line AB is divided externally at the point. /"
into the two segments P'A and P'B.
To divide a given line externally id a given ralin (us 2 : T), divide it
ratio (bkT' — 2, (ir .) fiarts) uiid ]iiMdiu;e it lill liic pmdutfd pait ciiuals
tlie siualler ti.rm of llie ratiu ihiK^ ilie unit line IVmud.
335. A line divided harmonically is a line divided Imih
inlprnally aud exUTiuilly iu the same ratio.
Thus, if the line AB is divided internally, so that PA .-
PB-l : 2, and externally, so that FA -. /"/>'=! ; 2, then
PA : FB = PA PB, and the line is divided harmonieally
Ex. 1. Divide a givea line barmonicallv in the ratio I ; 3.
Rs. 2, Divide a given line liarniaiiiciilly in ilio ratio ^^ : j.
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rKOPORTIOKAL LINES JLH)
Proposition XXII. Theokem
386. In any triangle, the hisector of an exterior angle
diriihs rxiernalhj the opposite side prodiiird into segments
which arc proportional to the uther two sides.
Given the A AUG with its sxtei-Lor Z UAO bisected by
le line AD ineetius tbe opposite side produced at D.
To prove
1)B: DC^AB: AC.
Proof. I
ru\
- the line BF II AD and meeting
AC at F.
Tliyii
DB: OC^AF: AC.
(Why!)
But
Z/' = Z/.
fWbyr)
Aud
Z. = Z/..
(Whj!)
Also
lp'=Zp.
.: Zi-= Is.
(Why?)
(Why!)
A
ADFia isoseeles, and AB
= AF.
{Why})
Snbstiti
: AB for its e(inal, AF,
11 the
al)
ve ]>vo-
ortiou.
DB -. DC^AB : AC.
i\\. S.
337. Cob, The lines bisectttxj the iiUn-ior and ixtcrhr
anf/lis of a Iriaii'jk at a given vertex, divide- the opiiusite
siiJt karmoiiicully.
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Proposition" XXIII. Theorem
338. The homologous altitudes of two similar ti-ian'jlet
have the same ratio as any two homologous sides.
Given the similai- A ABC, A'B'C, und IW, B'J)'
any two homologous altitudes in these ife.
B I) B A BO A a
To prove WD' = WA=Wo'^Tr/
Proof. In the rifrht A ABD and A'H'iy,
Hij hiiiimhigiius A of tbesiniilfir & JlWanil A'B'<'').
:. A ABI) aiul A'B'I)' are similar, Art. -.i-^A,
A hace an acuta L of one — an acute I of the other, the &
■• B'jy B'A'' ''^''^' ■'"'■
1 the similai' A ABC and A'l.i'C,
BA BC A C
Wa'^Wc'^a^' ^"■^^'■
b i> ^ ba ^bc ^a c _
b'd' b'a' b'c a'c' ^"^ ^'
<). E. ».
(, AC=18, A'C = 12, and tlD =
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I-ROPORTIONAL LINES 201
Proposition XXIV. Theorem
339. If three or more lines pass through the same point
and intersect two parallel lines, they intercept proportional
segments on the parallel lines.
Given the transversals 0^ , OB, OC, 01) intei-secttnE the
parallel lines Ali and A'D' in the points A, B, C, D and
A'. B', 6", !>', respectively,
Ali B a (J n
To prove a1^'=JF(J'^c1^-
Proof. A'D' 11 .1 P. R.rp.
Therefore the hiise A. of the A A'li'O, B'C'O, et,e., are
equal to corresponding base A of the ^ ABO, BCO, etc.
All, nti <n- Art. [2.1.
Hence the A A'B'O, B'C'O, etc, are similar to the A
ABO, BCO, etc., respectively. Art. 324.
iVB' Kii'o) B'V \V'Oj C'l)' ^ '■■
Es. If tbe skies of a polyt,'oii
111* polygon, the side homologous
the second polygon.
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MJ, BOOK III. I'LASE GEOJIKTllV
Prop. SXV. Theokem {<Jonvekse of Prop. XXIV)
340. // three or more non-parallel straight Unes inter-
ccpi proportioHul si'/jmrnts oh two pnraUds IJiey pass through
the name point [thiiC -is, are coiu-iirreiil).
psirallel lines AC and A'C f
that -
Given the transversals AA', BB\ CC intersecting the'
AB ^ BC
' A'B' B'O'
To prove that the lines AA' , BB' , CC are eoiicurreut.
Proof. Pt'odnoe the tines AA' and BB' to meet at some
point 0.
and let it intersect t
)nnv the lir
B point r.
A'V Al
1 H
IW
But
HeDue
A'B' B'P
A'B' B'C
B'1'=B'C'. Art, ;
.'. point P coincides with point C.
.: line CC coincides with line CP. Art.
:. the line CC, it produced, passes through 0,
(for it coincides with the line CP, iBliich passes through 0).
:. AA', BB', CC meet iu 0.
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PEOPOBTIOXAL LlXh;s 203
Proposition XXVI. Theohem
341. The perimeters of two similar pohjijotis have ike
same ratio as any two homologous sides.
GiveE t Iw two Jiimiliir i">lyt,^uLi.s A IIUDE and A'/i'VITI!',
with their perimeters denoted by P and P' Slid with AB aad
.I'/i" itny two homologous aides.
To prove that F -. r = AB : A'li'.
Proof, AH A'H'^BC : li'V^CD : G'I)'^<-t<i. Art. 321.
Hence An+IW+Cn+,i-tv.. : A'fr+ B'V'+V'I>'+,etc.,
■^Ali : A'B\ Art. 312.
(iii a wnra of rt/ml rirfio.'^. Ilin s,:,„ <,/ ,<n liir t,:lri-e<lenl>! !« /" Ihrn^im „J
<\ll Ihc eomequcnls as amj m,t ,u,UT.cilcnl is to its cons<:>pieuf).
Or P:/" = -t/i ; A' !S' .
Q. E, D.
341 (CI. Col;. !i, tiro :<i mile, 'j.ohii/on--^. uiiij two huuwl-
iKjoax lini-i air hi rack other as iinij other lii'o hoiiioLoijoiis liutx;
and the (xriiwter^ are to each other asainj two homologous lii,(n.
Ex. If the porimuUT of a dv.'ii li.'ld is 210 ridf^ ami i. siilr «t ihia
Held is ti. !L himioloHdus Ad<- of a similar li.Od iii ^ ; 2. tioil Uie i>crmie-
ter of the iouiucl litlil.
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Proposition XXVll. Theorem
342. In a right trianijle,
I. The altitude to the hypotenuse is a mean proportional
hetween Die segments of the hypotenuse;
II. Each leg is u mean proportional between the hypote-
nuse and the segment of the hypotenuse adjacent to the given
leg.
Given thfi Hglil. A ABO and BF ihe altitude upon the
hypoteniiMe A G.
To prove !. .IF: HF^HF-. FC.
I A<! : An = AH: AF
\ AC : BC^ liC: FC.
Proof. Th(! Z A is coiriiiinn to the rt. A -4 BF and ,4 BC.
.: A ABF and ABC are similar. Art. :i*J4.
Similarly Z <7 is common to the rt. A BFC and ABC,
and A BFC and ABC are similar.
.-. &. ABF" and BFC are similar, Art. 3-25.
{if two & are similar to the sam« A, they are similar to each other).
Hence, I. In tlie A ABF and BFC,
AF : BF-= BF : FC, Art. a21.
{homuhfjin'f suies of simitar & an: pruporUonal),
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PROPORTIONAL LIXES "JUS
n. In the simiiar ^ ABC uiid ABF,
AC : AB = AB : AP. Art. 321.
Also, ill the similar A ABC and BFC,
AC:BC=BC: FC. Art. 321,
Q. E. B.
343. Cor. The pet-pendicular to the diameter from any
point in the eircumfercnce of a circle is
poriional ieiween the segmeiiis of the di-
ameter; and the chord joining the point to
an extremity of the diameter is ,a i
proportional heltveen the diameter and
the aegmcnt of the diameter adjacent to the chord.
344. Def. Tbe projection of a point upon a line is the
foot of the perpeudicular drawn from the point to the line.
Thus, if AA' he perpendicular to LM, A' ia tiie projec-
tion of the jHiint A on the line LM.
345. The projection of a line upon another given line is
that part of the second line which is included between per-
pendiculars drawn from the extremities of the first line
upon the second line. Thus, the projection of AB on LM
isA'B'; of CD on PQ. is CT).
Ea. 1. IE the si!|-mentB of tho b.V[ioteimae of a right triangle are 3
and 12, find the ultitiide on tlie hypotenuse ; also find tha legs of the
triangle.
Ex. 2. If, in the figure of Art. ^43, the ciiainetet is 20 and the longer
chord IK, ti(]ii the hegiiicnt of lliu diu.mets;r udjucent to the chord.
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Pkopo-^itiox XXVIII. Theorem
346. In a rigid inawjli-, the square of the hypol'-na^i
equal to the sum of tlir K(;'(«ivx of the Imjs.
Given the right trl
,IC thi' liypot^TiuPfl.
Proof. Draw the line BF ± AC. Tlieo
AC : AB^Ali : AF :.AC X AF= AI?. Arts
Also AC: nC=nC : F<: .-. AC X FC=BC^.
Adding equals, At' (AF-+ FC) ^Ti? -rluf .
Or 'aJ^ = AJr + luf.
347. Cor. 1. 7n a right triang!'', the square of '■ilher
leg U equal to the sqi^are of the hi/poti-nuae mliii<.s the aquari'
of the other leg.
348. Cor. 2. In the square ADCf>. the diagonal di-
vides the square into two right ti-iangles.
Henee AC- = AB' + 'bC'=2 AJT.
.-. AC^^ABV2, '>r~ = '~-
Hence flte diagonal and the side of a
square are incommtnsurahie.
Ex. 1. If the legs of a
Ei. 2. In tiie figure o
t. A are 1 J- in. and a in., liiid the liypotenuaB.
p. 204, show that ~AB- : BC^ = dF : FC.
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_Vi;MEIiICAL fROlEiri'IKS -07
Proposition XXIX. Theorem
849. In any oblique triangle, tlte square of a side oppo-
xUp an acute angle is equal to the sum of the sguares of the
other two sides, diminished by twice the product of one of
those sides hj the projection of the other side upon U.
Given neiite ^ (7 in A ABC, and DC tlie iii-ojet
the side BC on the side A C.
To prove lB-=BC^+AXf — 2 AC X DC.
Proof. If P fulls oil AC (Fi^. 1), AI)==AC—r)C.
If Zt falls on ilCpi-odueed (Pig. 2), AD^DC—AC.
In either case, Air^AC^ + IHf- — 2 ACX DC. ax. i.
Adding Blf to each of these equals,
I7r + £Z*'=l6'' + 7*t'' + !B7r— 24CX dc.k^.i.
But, in thei-t. A ABD, Air + 1^)^ = ~AB" , An, -.m,
and, in the rt. ABBCIKr+Bir^'BC^. (Wbyf)
Substituting these values in the above equality.
AB^ = mf + ACP — 2 AC X BC. ax, 8.
1, A line 10 i
ind tlio projeetioi
Q. E. K.
Ex.
line; i'
[1. loii« ninkes on nnsle of 45" with a seeond
;i of tli6 first line on the senoiid
Ex.
2. Find the 9a
me, if the angle i» tiO°.
tion ot
3, If the aide i
1 the base.
jf an equilateral triivngle is a. find it'f projeo-
Sx.
4. If, in I'ig. 1
, iX'=10, AC = 1':, ttud 6C = ^d% lliid.-!«.
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PliOl^UWlTION XXX.
350. In any obtuse triangle, tJie square of the side oppo-
site OH ohtuse angle is equal to the sum of the squares of ike
other two sides, increased hy twice the product of one of the
' ' by the projection of the other side «j)oh that side.
sides
Given the obtuse lAGB iu the A ABC, and CJ) the
projection of BG on AC produced.
To prove AB^= Jo*+"EG^+ 2 AC X CD.
Proof. AD=AC+ CD. As. 6
.-. A3'='Ad'+'CB^+2ACXC}). A^. i.
Adding BI> to each of these equals,
AI)~-\-inf=^^+'CJJr+'BD^+ 2 ACX CD. Ax. 2.
But, ill the rt. A ABD, AI?+ lw^= AB'-. (Why f)
And, ill the rt. A CBD, CI?+ 'bd'= 'BCT. (WLy •<)
Substituting these values in the above equality,
'A}?'='A<f+'B<f-\-2 ACY. CD. Ax. 8.
Q. Z. S.
361 . Cor. // the square on one side of a tnangle equals
the sum of the squares ott the other two sides, the angle oppo-
site the first side is a right angle; for it cannot be aeute
(Art. 349), or obtuse (Art. 350).
Et. 1. If, in the above ."igure, BC=10, AC = 2, and IB€A = 120'',
find AB.
El. 2. UAB = 2I}, ISC=U, iind AC = 12, find CD.
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NTiMERIGAL I'EOPEItTlES 209
Proposition SXXI. Theorem
352. J/, in any triangle, a median be drawn to one side,
I, The sum of the squares of the other two sides is egual
to twice iJie square of half the given side, increased hj twice
the square of the viedian upon that side ; and
II. The difference of the squares of the other two sides is
equal to twice the product of the given side by the projection
of the median upon that side.
Given the A ABC, AB>AO, AM tlie median upon BC,
and MF the projection of AM on BC.
To prove I. Air+ lc^=2 32?+ 2 'AM^.
11. Is — Ic^=2 BC X MF.
Proof. In the A BMA and AMC, BM^SIG. AM^AM,
&ndAB>AC. Hjp.
.-. ZAMB k gresiter thun /-AMC. Art. 108.
,". ZAMB is obtuse {for His greater ilian half a straight Z),
luohtme £S ABM, A^ =
In acute A ACM, AG^.^MO^+AM^— 2 .
Adding, Jb'+Ic'=2 BM''+ 2 AM'',
{forMC=im).
Subtracting, Tlf-^AX? = 2BCX MF.
Ifor BM+MC=BC).
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PLANE G eOJlETRY
353. Formula for median of a tiuangle in terms of its
sides. Ill the Fig., p. 209, deiiothij; AB by c, AC by b,
BG by a, and AM by )», by Art. 352, b'^ -\- 'j^ ^ 2 m^
-K?)%
'■^(/y^ + .
354. If iu-o chords in i
the segments uf one chord i
inmts of the other chord.
XXXII. TUEOEKM
a circle intersect, the product of
rqiial to the product of the seg-
Given the O ADBC with Uie chords AB and CD inter-
eectics at the point F.
To prove AF X FB= OF X FJ).
Proof.
Dm
y AD and GB.
Then,
in the & AFn ami CFB. lA- IC.
(mk/i hcinfj jNeHSMra? hij i it:-/: IHI).
Art. 2r,».
Also
II)=IB.
(Wlyt)
Hcuoe
the
& AFD and C FB are similar.
(Why f)
.-. AF: CF.FD: FB.
{Why I)
And
AFXFB=CFX FD.
(Wbjt)
Q. £. S.
855. Cor. 1. If through a fixed point within a circle a
chord be drawn, the product of the segments of the chord is
constant, in whatever direction the chord be drawn.
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PROPORTIONAL LINES 211
356. Def. Four directly proportional quantities are foar
quantities in proportion iu such a way tliat both, the ante-
cedents, bolong to, one figure and both the consequents to
another figure.
Four reciprocally proportional quantities are four quanti-
ties in proportion in such a way that the means belong to
one figure and the extremes to another figure.
357. Cob. 2. The, segments of two chords intersecting
in a circle are reciprociUy proporlional (tiie segments be-
ing considered us parts of the chords, not of the A).
Proposition XXSIII. Theorem
358. If, front a given point, a secant and a tangent be
drawn to a circle, the tangent is the mean proportional he-
tween the whole secant and its external segment.
Given AB, a tangent, and AC, a secant, to the cLrele
BCF, and AF Vad external segment of the secant.
To prove AG : AB^ AB -. AF.
Proof. Draw the ehords BC and BF.
Then, iu the A ABC and ABF, Isi ^ lA. Ident.
IC= I ABF. Arts. 258, 264.
.-. the ^ ABC and ABF are .'^iiiiihu-. (Why n
.■. AC : AB = AB : AF. (Why !)
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J-U HOOK III. PLANK GF.OMKTKY
SbQ. Cor. 1. If, from a given point, a tangent and a
secant he drawn to a circle, the product of the wJiole secant
and its external segment is equal to the square of the tangent.
860. Cor. 2. If, from a gi fen point without a circle, a
secant he drawn, the product of the secant mul its external
segment is constant, in witatever direction the secant be drawn.
Fop the product of each secant and its external seg-
ment eqnnls the square of the tang'ent, which is constant.
361. Cor. 3. If two secants be drairii from an exter-
nal point to a circle, the whole secants and their external
segments are reciprocally proportional.
pROPOPiTiox XXXIV. Theorem
362. The square of the bisector of an angle of a triangle
is equal to the, product of the sides forming the angle, dimin-
ished by the product of the segments of the third side formed
h'j the bisector.
Given the A ABC, CF {or t) the bisector of ZACB,
aad m and n the segments of AB formed by CF.
To prove t^ — ab — nin.
Circumscribe a © about the A ABC,
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KUMEKICAL PISOT'EETIEa
Proof. Prbduee OF to meet the circumference
at H",.aud
di-aw the chord BH.
Then, iu the A ACF mdi CHB, AA = III.
(Why ?)
And IACF= IHCn.
(Why !)
Hence the A ACF and CUB are similar.
(Wby?)
.-. 6: £ = ic + (:a.
(Why ?)
.-. ( {xArt)=al.
(Why!)
Or (a- + /2 = ai.
Subtracting ix, f = (ih — tx.
Ax. 3.
But /;c = WH.
Art. 3r)4.
Substituting liin for te,
363. Formula for bisector of an angle of a triangle.
m:n = h :«(Art.332).-.m + K:m = i + (( ; (.{Art. 309),
or c ; m — o■■^- h -. h (Ax. 8^
he
In like manner K^^ — —r- Substituting for i
(I + b
nhr^ _oh(f< +h + r){a +h-
(^=--ab-
Ei. 1. Oil the figure, p. 210, lot ^F=10, rR = 4, and FC^S.
'mil FI>.
Ex. 2. On thfi flfc-urH, p. 211, let AC = IG ui.d .JC = ia, Find AF.
Ex. 3. On the figuvo, p. 213, let 4C=1G, CB = 12, and 4C = 14.
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Zli JiOOK in. PLANE GEOMETHY
Pkopositiox XXXV. Tuboreji
364. Jn any triauyle, the product of any two sides is
equal to the product of the diameter of the circumscribed cir-
cle by the altiiitde upon the third side.
Given the A ABC, ABCJ) a circumscribed O, BDthe
diameter of this O, and Bi^'the altitude upon AC.
To prove AR X BC=BD X BF.
Proof. Draw the chord DC.
Then, in the A ABF and DBG, I A = ID. CWhyS)
IDCB is art. Z. (Whyf)
.-. & ATJJi'and 7>/J6'are simihir. (Why!)
.-. AB : BB^BF: BG. (Wl.yf)
.-. A B X BC=BD X BF. (Why?)
q. E. D.
365. Cor. The diameter of a circle circmnscribed about
a triangle eqiiah the product of iivo sides of the triangle
divided by the altitude upon the third side.
Ex. In the above figure, if AB = 8, CC = U, and ilF = 4^, fled the
rtidiua o£ tbe eireumBcribed (.'iicle.
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construction problems 215
cohstroction problems
Proposition XSXVI, Problem
6. To construct a fcurth proportional to three given
Given the lines m, n, p.
To construct a fourth proportional to «s, n, iuid p.
Construction. Take auy two liuea, .4/* and AQ, making
any convenient /.A.
On J-Ptake J.B = »k, and BC^n.
On AQ take AI)= p.
Draw BB *
Through the point C draw CB\\J)n, and meeting A^
at E. Art. 2T9.
Then I>Ii is the fourth proportional required.
Proof. An : BC=AI> : DR. Avt. :siT.
Or m : n^p : DU. Ax, S.
Proposition XXXVII. Problem
367. To consiriiet a third proportional to two given lines.
Let the pupi! supply the construetion and pr'oof .
[SUQ. Use the method of Art. :i6C, ninking p = n.]
Ex. 1. Construct a fourth proportional to three lioea, j, 1 and
I In. long.
Ex, 2. CocBtnict a third proportional to two lines, 2 and 14 in. long.
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BOOK m. PLANE GEOJIETliif
Proposition SXXVIII. Problem
To construct a mean proportional between iivo given
Given the lines m and n.
To construct a mean proportional between m anil n.
Construction. On the line AP take AB= m, and BC=
At B erect a ± to AG meeting the semi-eironinferenee
at R. Art. 274.
Then BE is the mean proportional reqnired.
Proof. AB : BE=BB : BG, Art. 343.
{Ihc J- to the diameter from atiij points in the eirciniferance of a circle is
a mean proportioiial belweeii the segmenls of the diamvler).
Substitnting for AB and BG their values »« and n,
m : BB^BK -. n. Ax. 8.
Q. E. F.
Ex. 1. Construct the third propovtloaal to two I'mes, 1 and IJ in.
long.
Ex. 2. Construct a mean proportional between two lines, 1 and 2
in. long.
Bi, 3. Taking any line as 1, oonstiniot j/S.
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CONSTRUCTION fROELEMS 2l7
Peoposition XXXIX. Peorlem
369. To divide a givm straight line into parts propor-
tional to a number of given liiies.
"^'
\
Given the straight lines AB, m, n aiiJ /!.
To divide AB into parts proportional to »t, n and j).
Construction. Draw the line AP, making any convenient
angle with AB.
On APmark off AC=»H, CD=n,mdI)F=p. Draw-BP.
Throngh the points C and D draw lines || BF, and meet-
ing AB at E and 8. Art. 279,
Tliou AR, US and SB are the segments required.
_ , AB ES SB
^"°^- AC^cB^W ^''■'''-
{if two lines are cut by a wimbey ofparallets, the corresponding segments
are iiroportional) .
For A C, CD and BF substitute their equals »», n and p.
„, AB B8 SB
Then — = — = —-. ax. h,
m 11 2)
Q. E, If.
370. Def. a straight line divided in extreme and mean
ratio is a straight line divided into two segments such that
one of the segments is a moan proportional between the whole
line and tiie other segment.
Ea, Divide a given line lato parts prupc
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18 BOOK III. PLANE GEOMETRY
Proposition XL. Problem
371. To divide a given straiglii line in extreme and,
Q 'I
Given the line AB.
To divide AB in extreme and mean ratio.
Construction. At one end of the given line, as B, con
struct a X OB eqnul to \AB. Arta. 274, 275,
From as a center and with OB as a radius, describe
Sl circumference.
Draw AO meeting the circumference at C, and produce
it to meet the circumference again at F.
On A B mark off AP eqnal to A€\ on BA produced
take A<J = AF.
Then AB is divided in extreme and mean ratio inter-
nally at P, and externally at Q.
Proof. 1. AF: AB=-AB : AC. ah. n5fi.
.-. AF~An:AB=AB — AG:AG. (Win-?)
But CF=20B = AB, and AG=AP.
Hence AF~AB=AP, and AB — AC^PB.
.: AP:AB = P£':AP,orAB:AP=AP: PB.
As. 8, Art. 308,
2. AF: AB = AB : AC. (Why?)
.-. AF+AB:AF=AB + AC:AB. (Why!)
But AF+AB=QB. anH AB + AC^AF^QA.
:. QB: QA = QA : AB. Ax. B.
Q. £. F»
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consteuction problems 219
Proposition XLI. Problem
S72. Upon a given straight line, to construct a polygon
similar to a given polygon, and similarly placed.
Given the polygon ABODE and tlie line A'B'.
To construct on A'B' a polygon similar to ABODE and
similarly placud.
Construction. In the given polygon, draw the diagonals
AG and AD, dividing the polygon into triangles.
At /i'^on the line A'/f, eonstruet ZA'B'O' equal to /.B;
and at A' construct Zll'A'O" eqm\\ to ZBAO. Art. 2TS.
Pi-oauce the lines B'C and A'C to raeiit at C.
In like manner, on A'O' construct A A'C'D', equiangular
with A AOD and similarly placed; au.l on A'D' construct
the A A'D'E', equiangular with A ADE uud similarly
placed.
Then A'B'C'D'E' is the polygon required.
Proof. The ^ A'B'C, A'C'D', etc., are similar to the
& ABC, ACD, etc, respectively. Art. n24.
Hence the polygons A'B'C l>' l-l' and ABCBE are
similar. Art. :iJO,
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PLANE GEOMETKV
EXERCISES. CROUP SO
SDllLAR TiilANGLES
Let the pupil make a list of all the conditions that ?
tivo triangles similar (see Arts. 323, 324, 325, etc).
Ex. 1. Given AD ± BC, and BF 1. AC; prove
& ADC and BFG similar.
Ex. 2. In the samG figure, prove the
and BFC similar. What other triangle
figare i3 similar to A BFC f
Ex. 3. Two isosceles triangles are si
theii
uqual
Ex. 4. Two iBOBGeles triangles are Bimi
equals a base angle of the other.
Es. 5. Given arc -iC=aro J!C; prove A
AFC and AFC similar.
Ex. 6, Prove that the diagonals and bases
of a trapezoid together form a pair of similar
triangles.
Ex. 7. AB is the diameter of a circle, BD is
a tangent, and AD intersects the clreumfetenco
angles ABE and ADB similar.
Ex. 8. liC is a chord in a circle, AQ is the di.imeter perpendicular
to BC and meeting it at 2f; AF is any chord intersecting BC in M.
Pi'ove the A JJfif and APQ similar.
Ex, 9. The triangle ABC is inscritied in a circle; the bisector of
the angle A meets BC in 1) and the circumference in F. Prove the
triangles BAV and AFC similar.
Ex. 10. A pair of homologous medians divide two similar triangles
into triangles which are similar each to each.
Ex. 11. Two rectangles are similar if two adjacent sides of one
are proportional to the homologous sides of the otlier.
Ex. 12. Two circles intersect in the points A and B. AC and AD
are each a tangent in one circle and a chord in the other. Prove the
iiJ^C and JBi) similar.
[8ufl. Prove IBAD= ^ACB, tti!.]
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EXEBCISES. PROPORTIONAL LINES 221
373. Proof that lines are proportional. In order to
prove that certain lines are proportional, or have propor-
tional relations, it is usually best to show that the given
lines are homologous sides of similar triangles.
Sometimes, however, other methods of proof are used
(as the theorems of Arts. 354 and 358) ; but these, if in-
vestigated, are usually found to be the method of similar
triangles in disguise.
EXERCISES. CROUP 31
PROPORTIONAL LINES
Ex. 1. Ontho iignpe of Ex, 1, p. 220, prove JI'X /JC=B-FX--ir',
and TSC'X Or> = BOXFC.
Ex. 2. On the figure of F.k. 5, p. 320, prote CP: CA=CA : CF.
(Hence as P moves tho product of what two lines is constant i )
Ex. 3. The diagonals of a trapezoid divide each other into pro-
portional eegments.
Ex. 4. In the isosceles triangle ABC, AJi = AC,,oa tho side ^fl,
the point P is taken so that PC equals the base. Prove AB X PB = m?.
Ex. 5. In a triangle the median to the base bisects all lines
parullel to the base and terminated by tlie sides.
Ex. 6. If PQ is nnj- line throngli F. tl.e midpoint of the line AB,
and AI' mid BQ are perpendicular to PQ, show thai the ratio PF: FQ
Ex, 7. The triangle ^CC is inscribed in a circle. J^is the midpoint
of the arc JC and Bi*" intersects the line 46' in ffi Prove AB : liO =
AE: EG.
[SUG. I'se Art. 332,]
Ex. 8. If two circles intersect, the common ehoid, if produced,
bisects the common tangent.
[SUG. Use Art. 358.]
Ex. 9. If two circles intersect, tangents drawn to the two circles
from any point iu the common ishord produced ure equal.
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22'2 BOOK III. PLANE GEOMETEY
Ex. 10. Givi'Ti AJ>. I'T and EG I ;
prove PQ = UT.
[SuG. Show that ^= If', by show-
ing them equal to a uommon vatio.]
Ex. II. Lilies aveiltawn from a point within the triangle, to the
vertices of llis triangle AltC. From B' any point in OB, £'A' is
drawn parallel to J)'J and meeting OA in A', aud B'C' is drawn parallel
to BC and meeting OC in C Prove A'li' : AB=B'C' : BC, and the
ti'iangles ABC and A'D'ff similar.
Ex. 12. GiyenJ7(c;Dn£I7,and ^ ^
F any point in BC produced | prove
AE' = IIQXI1P-
[St'G. Compare the similar A
ABR and BQD ^ also the similar &
AMD and EBP.'] " "
eXEFtCISES. GROUP 85
NUMF.RICAL PROPERTIES OP LINES
Ex. 1. If AD is the altitude of the triangle ABC, Tl^—lc?^
luP'—m:"- .
Ex. 2. If the liingoiials of a quadrilateral are perpendicular to
each other, the sura of the squ.ares of one pair of opposite sides
equals the sum of the squares of the other pair of aides.
Ex. 3. The square of the altitvido of an equilateral triangle is
three-fourths the square of one side.
Ex. 4. If AB is the hj-potenusc of a right triangle, and the leg BG
is hisected at K, ZZJ" - AM? = SOff-.
Ex, 5, PlJ isaline parallel totbehypotenuse^J of arighttriangle
^JJC, and meeting JCiuPand^Cine, Prove Z^^ + 5P = Zb^+P^2.
Ex. 6. Ill the right triangle AhC, BE and Cf bisect the legs AG
naAdB in the points £ and F. Pvove iR^ + ici^=5B&.
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PJ^-
-PC^
EXERCISES. AUXILIARY LINES
EXERCISES. CROUP SS
ArXILIARY LIKES
iiven ^Br/i ft veutHnglo; prove
= 'PI'?-\-'PD-.
Ex. 2. The
eles divides tlie line of centers into segments
which have the same ratio as the diameters
ol the oircleE.
[Suo. Draw radii to the points of coi
Ex.3.
n altitiidi
■IB and -■IC are the liigs of a
Prove 2JCX-F'C = i(6'2.
Ex. 4. Given the chords AB and CD perpen-
dicular to each other and intersecting at O; prove
(W^ + OB^ -r OC^ + 0O^= (diameter)^,
[SuG. Draw tlie diameter BE and the chords
AC, BD, DE. Prove AC=ED. etc]
Ex, 5, ABC is an inacrihed i
of which AB and AC are the legsi
Ftov6A^=ADXAE.
AD is a chord r
3elos triangle ABC, and D is
n the t
6 produced, then CD^=CB- + ADX BI).
Ex. 7. Two circles touch at the point T. I'TI" and QTO" are lit
drawn meeting the circumferences in P, Q and P", (/ resiiective
Prove the triangles PTQ and F'TQ' similar,
[Sua. Draw the common tangent at T.]
Ex. 8. Two circles touch at the point T, throngh 3" three lines i
drawn meeting the circumferences in i', Q, E and P', (/, E', reapi
tively. Prove the triangles PQR and PQ'Ji' similar.
Ex. 9, If A is the midpoint of CD. an arc of a circle, and AP
any chord intersecting the chord CV In Q, prove that APKAQ h
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2^i BOOK ill. PLANE GEOMET
Ex. 10. In nn iiiseribed qnadriluterdl, the
product of the dingonala ia equal to the sum o£
the products of the opposite sides.
[8UG. Drfiw BFfO that ICBF= lABD and
use similar trian{|;lts.]
Ex. 11. The sum of the siiiian's of t]iy
aides of any quadrilateral is equal to the sum
of the squares of the diagonals, plus four times
the square of the liue joining the loidpoiuta of
the diagonals,
EXERCISES. CROUP 34
INDIRECT DEJIOXSTRATIONS
Ex. 1. If the sum of the squares on two sides of a triangle is
greater than the square on the third side, the angle included by the
two given sides is an aeute angle.
Ex. 2. If D ia a point in the side AC of the triangle AJ!C. ajid
AD : DC = AB : BC, then Dli biseeta angle ABC.
Ex. 3. A given stmight line can be divided in a gtveu rritio at but
one point.
Ex. 4. If the sides of two triangles are par-
allel, each to eauh, and a straight line be paasiid
through each pair of homologous vertices, these
lines, if produced, will meet
Ex. 5. If each of three c:
other two, the three common chords intersect in
EXERCISES. CROUP 35
THEOREMS PROVED BY VARIOUS METHODS
Ex. 1, In the figure on p, 204, show that ABy, BF=BCX AF.
Ex. 2. In the same figure, if J'C=3vli'', sliowthatZS^ : fl(?- = ! : 3.
Ex. 3. AB is the diameter of a circle and PB is a tangent. If AP
meets the ciccumfeteneo in the point §, prove that APX. AQ = AB\
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MIfiCELI.ANEOTIS EXERCISES. THEOREMS
Ex. 4i. If the line hisocting the parallel sides of a tra,pezoid b
produced, it meets the legs produced in a common point.
[SUG. See Art, 340.]
Ex. B. In similar triangles, homologous uiediaua have the aam
ratio as iiomologoua sides.
Ex. 6. A diameter AB is produced to the point C : CI' is perper
dicular to JC; PB produced meets the eiruumterence at Q. Prov
the triangles A QB ai.d POP. similar.
Ex. 7. If PA and PB are chords in a eirele, and CD is a line pai
aliel to the tangent at J* and meeting PA and
PJl at C and I), the triangles PAIS and PCD are
similar.
Ex. S. Given AB a diameter and AD and BO
tangents, JCand DB intersecting ut any point F
on the circumference; prove AB a mean propor-
tional between the aides AD and BC.
Ex. B. In any isosoelea triangle, the square
of one of the legs equals the square on a line
drawn from the vertex to any point of the base
plus the product of the segments of the base.
Ex. 10. A line drawn through the intersection of the diagonals of
a trapezoid parallel to the bases and terminated by tho legs is bisected
by the diagonals.
[SuG. SeeEs. 10, p. 222.]
Ex. 11, If a chord is bisected by another ehorl eath segmi
the first chord is a mean propoitiouil letweei the begui'nts of th
second chord.
of
Ex. 13. If two circle'* aie tangent exteinallj and a linn is di:
through the point of lunta t terminated bv the cireumferoncea
ohordu intercepted in the two circles are to ^ath other ai the rain
Ex. 14. Thz-ee times the sum of the squares of the sides oE a
Bugle equals (our times the sum of the squares of the medians.
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2'^(>
BOOK ni. PrANK
El. 15. Find the loeua of the jHklpoiuts of
lines in a trinngle parallel to the biis" and ter-
minated by the sides.
Ex. 16. Given AB the iliam^ai-r. AP. PQIi,
BR, tangents; pmve P^ X Qli a<:ousUnt(= ra-
dius squared j,
Ex. 17. is the center of h circle and A is
any point within the circle ; OA is produced to J),
so that OA'KOB equals the radius squared.
If P is any point in the oivcumferonee, the
angles 0J>^ and OBI* are equal,
[SUG. Use Art. 327.]
Ex. 18. Given AF=FI1, and ri{ \\ Jl! :
prove BP : FP=HK : FK.
[SuQ. EP : FP=UII: FH, utc]
Ex. 19. If from any point P within the triangle AJIC the perpen-
diculars /'§, ^.B, PTare drawn to the sides Jii,JC, BC, re^ p. actively,
EXERCISES. QROUP 3a
Ex. 1. Construct J- = — ; ulso .c= — .
Ex, 2. Construct i- = \/ <i'— iF~, i. a., l/'lTT'f Th"-^.
Ex. 3. Construct i = V''iai, i. B., "|/(3 u) /i.
Ex. 4. Construct ■x = \/d' — t>r', i.e., 1^ «" - {l/fiTo'.
Ex. B. Given a line denoted by 1, eoustiuct i/3; also iv/5.
Ex. 6. Divide a line inta three parts proportional to 2, \, %,
Ex. 7. Divide a line harmouieaily in the ratio 3 : 5,
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KXEKOISES. PROBLEMS !2'i7
ti'iangie into segments propoi'tioual la
Ex. 9, Divide aline into segmenta in the ratio 1 ; -j/S.
Ek, 10. Given a point P in the aide JB o£ a triangle ARC, draw
a line from P to AC produeeii so that the line drawu niiiy be bisected
byBC.
[SUQ. Suppose the required line, Pqp., drawn meeting liC in Q
andJCinB. From P draw Pi H JC. Compare the APL^ and yjiC]
Ex. II. Through a given point F in the are
subtended by the chord AB draw a ohord which
shall be bisected by AB.
[Suo. Suppose the required chord drawn, viz.,
PQR. Jointlieeeoter O with P and y. What kind
of an angle is OQP, etc,?]
Ex. 12. Ill an obtuse triangle draw a line from the vertex of ths
obtuse angle to the opposite Bide which shall be a mean pioporticmal
between the segments of the opposite side.
[Sua. Cireumscribe a eirole about the triangle and reduce the
problem to the preceding Ex.]
Ex. 13. Find a point F in the arc subtended by the chord AB
such that chord PA : chord PB = 2 : 3.
[SuG. Suppose the required construction made, and also the ehnrd
.^B divided in the ratio 2 ;3 at the point Q. How do the angles Ji'g
and QPB compared]
Ex. 14. Given the perimeter, conatruet a triangle similar lo a
given triangle.
Ex. 15. Given tliH altitude of a triangle, construct a triangle
Ex. 16. In a givun circle inscribe a triangle similar to a given
Ex. 17. About a given circle circumscribe a triangle similar to a
given triangle.
Ex. J8. By drawing a line parallel to one of the sides of a given
rectiiugle, dividt thu rcclanaio imu Iwo ainjilar rectangles.
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228
BOOK Hi. PLANE (lEOMETUY
Ex. 19. InKoribfl a square in a given
triangle.
[Si'O. If ABC is the given tnaagle, sup-
pose hOFE the required inaeribtd square,
.ioin HE and produce it to meet AH \\ IIC.
VroveJH=JK, etc.]
374. The metJiod of similars in solving geometrical
problems is best shown by the aid of
Ex. In the side JSC of a triangle JHt
tlie perpendiculars from it to tho othBr
sides shall be in the ratio 3 : 1.
CoNsTRUCTiOH. At any point /' in J''
erect ft X PQ of any convenient length,
lu a direction ± All draw Jfy= i Ql'.
ioia.EP. V-coiaSiTaw tiT\\IlQ. Produce
AT to D. Then D is the required point.
Let the pupil supply tiio proof.
EXERCISES. QROUP 3?
PROBLEMS SOLVED BY METHOD OF SIJIILARS
Ex. 1. In one side of a triangle find n point such tliat the perpen-
diculars from it to the other two sides siial! be in the ratio «i : n.
Ex. 2. Find a point the perpendiculars from which to tlio three
sides of a given triangle shall be in a given ri
[SUG. Use Ex. 1 twice.]
Ex. 3. Construct a oircle
which shall touch two givon
lin
thr.
given point.
[SUG. Let OA and OC bo
the given lines and P the
given point. Draw any OR toueliing the two liups {OB at r} and
intersecting OPprodueed at X. Dtaw the chord A'F, etc.]
Es. 4 Inscribe a. square in a given semi-circle.
[Sua. Circumscribe a semi -circle about any given square, by taking
the midpoint of the base of the square as a center, and the line from
this midpoint to a uou-adjaeent vertex as a radius, etc]
Ex. 5. Solve Ex. Ill, p 228, by the method of aimllara.
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KXEKCISl'lS . PROBLEMS
229
375. Algebraic analysis of problems. The cmuUtions of
a problem may ofitn be stated as an algebraic equation; by
sohiiig the equation, the length of a desired line in terms of
knotcH lines may then be obtained, and the problem solved
by construeting the algebraic e^^ression thiis obinined.
Ex. Find B point P in
the line AB mah tLat 'Jf-=
IBi-'. "
Analysis amd Construction. Denotu AB by ;
and PB\>ja — x. Tlien K' = 3{a — a;)'.
Construct a v'3, whanee construct '— — ; lay ofE the line ob-
tained, 88 AP, on AB; this gtv&s the point P of internal division.
Similarly, the eonatriiction of - ^ — gives F' , the point of external
3ISES. CROUP 3S
. SOI.VKD HY ALGI'IBBAIC ANALYSIS
E. tbatJ/'-=2BP-.
Ei. 1 . Find a point f in a given iii
En. 2. Construat a riglit triangle, given one i
a and the projection, 6, of the other k-g on i
[SuG. Denote the projection of a on the hy|
tennsebyai. Then o= = a: U' + 6), etc.]
Ex. 3, Inscribe a square in a given aemicircle.
Ex. 4p. From a given line out oft a part which shall he a mean
proportional between the rumaiuder of the line and another given line.
Ex. 5. Given AC and CB arcs of ilO°, and « a
given line. Draw the ehoiii CQ intersecting AH
in J'aothat7'y=(i.
[Sto. .c=-j' = y'. <r.i: = (r-i'i,) (/--J), elc] /
Ex. 6, Given the greater segment of a line di-
y\C -i extreme and mean ratio, construct the line.
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Zdii BOOK III. PLANE GEOMETTiY
EXERCISES. CROUP 3Q
PROBLEJIS SOLVED BY VARIOIIS .MF/I'HODS
Ex. 1. Conatruet two lines, given tljeir Hum [a line AB) and their
ratio {m ; «).
Ex. 2. Construct two lines, given tiioiv diffovencK iind tlieir ratio.
Ex. 8, Divide a trapeaoitt into two sitoilur tviifitKoids by drawing
a liue parallel to the bases.
[Suo. Coneeive the fisiire drawn, and compare the ratio of the
bases in the two trapezoids formed.]
Ex. 4. Construct a mean proportional between two given lines by
Ex. 6. Prom a given point draw a seeant to a eircle so (hat the
extprnal segment shall equal half the secant.
[SuG. Draw a tangent to the O and use the algebraic method.]
Ex. 7. From a given ©sternal point P, draw a secant meeting a
circle in A and B bo that FA : AB= m : n.
[SuG, Draw a tangent to the circle from the point P and denote
its length by *. Denote PA by inx and ^S by lij-. Then in{m-\-ii)x'
Ex. 8. Through a given point P draw a straight line so that the
parts of it, incliKied between that point and perpendiculars dvawn to
the line from two other given points, shall be in a given ratio.
[SuG. Join the last two points, and divide tlie line between ttiem
in the given ratio.]
Ex. 9. Conatruet a straight line so that the perpendioulacB on it
from three given points shall be in a given ratio.
[Sue. Let F, Q, R, be the given points and m : n : p the given
ratio. Divide FQ in the ratio m : n and Qlt in the ratio n : p, etc]
Ex. 10. Upon a given line as hypotenuse construct a right tri-
angle one leg of which shall be a mean proportional betwf - tlie
other leg and the hypotenuse.
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Book IV
AREAS OF POLYGONS
376. A uait of surface is a square whose side is a unit
of length, as a square inch, a square yard, or a square
etutirneter.
377. TIip area of a surface is the number of units of
surface which the given surface contains.
It is important for tbe student to grasp firmly the fact that area
menus not mere vague largeness of surface, but that it is a nainber.
Beinf^ a number, it can be resolved into faKtors, it may be determined
as a product of simpler numbers, and handled with ease and precision
378, Equivalent plane figures are plane figures having
pqunl areas.
Thus two triangles may have equal areas (he eqiiivaletit)
and ret not be of the same shape, that is, not be equal
(congruent).
379, Abbreviations. Instead of "area of a rectangle,"
for example, it Is often convenient to say simply "rectan-
gle." So instead of "the number of linear units in the
base," we use simply "the base." In like manner, for
"product of the number of linear utiits in the base by the
number of linear units in the altitude," a common ab-
breviation is "product of the base by the altitude."
(231)
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-•J-i BOOK lY. PI,ANE GEOMETRY
COMPARISON OF RECTANGLES
rROPOSiTiON I. Theorem
380. If Itco rectimijles have the same altitude, llinj <
to each other us their lases.
Given the rectangles EFGJT and J7>CI>, having their
altitiideri l-'Fand Ali equal.
To prove EFGB. -. ABCJ)^ hW -. A 7).
Case I. 'SVIien the hanes are (■ommensumNe.
Proof. Take some eomraon measure nf fT/f and .1/), as
AK, and let it be contained in EH n times and in Al) vi
times.
Hence EU -. AD = n im. (WLy?)
Through the points of division of the hases of the two
rectangles draw lines perpendicular to the bases.
These lines will divide EG into h, and AC into m small
rectangles, all equal. ^"- "'^'
Hence EFGR : ABCD^m m. {Wh;?)
.-. EFGS ■.ABCD = EU: AE. (Why!)
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COMPARISON OF RECTANGLES 233
Case II. When ike bases are incommensurable.
A D E Qil
Proof. Divide the base AD into Jiny numher of ftiiinL
parts, and apply one of these parts to EU.
It will be contained in EH a certain nnmber of times
■with a remainder QU, less than the unit of measure.
Draw gP ± EH, meeting FG at P.
Then EQ and AI> are commensurable. Oonstr.
. EFPQ _EQ
ABCD A3
If now the unit of measure be indefluiteSy diminis
the line QH, which is less than the unit of measure,
be indefinitely diminished.
.-. A-(>^£'/fu--alimit; EFPQ^^EFGHi>.i.i^\im\L An
EFPQ T^_,^„,„^ ^ „„,.ifl>,v wlch ^^^^ .
Case I.
381. Cor. // two rechinijUs have e(piul bases, iheij i
to each oiher ai> their ullilutks.
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264: ROOK IV. PLANK OEO^tF,TKy
PliOl'OSITION II. TilliUUKM
3S2. Tl)it ureas of any two reciani/hs are lo aii:h other
as till- proOiirta of their bases iij their altitudes.
Given the rectangles E and R', iiaving the bases b auiI
6', and the altitudes « and «', respectively,
_ R h X a
^'^''"'' R=VX7^'
Proof. Constrnet a rectangle, N, liaviug iUs ba^^e equal
to that of R, and its altitude equal to thai, of B'.
Also
R''
Taking the produet of the corresponding members of
the two equalities,
Q. E. D.
Ex. J. Fiud the ratio o
&re 12 X 8 iu. to that of o:
Ex. 2. How many brieks, eacL 8X5
avementeOX 9"-*
of a rectangle whose di
will it take to
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AKEAK OF POLYGONS JoO
AREAS OF POLYGONS
Proposition III. Theorem
3S3. The area of a rectangle is equal to the product of
Us base by its aUiluile,
,m
Given the reotanglefi, with a base containing b, and an
altitude containing h nnits of linear meaanre.
To prove area of B=b X h.
Proof. Let fTbe a square each side of whicli contains (
unit of linear measure.
Then t-'^is the unit of surface. ArL. 'sm.
1X1
= bXh.
— is the ar
aof B,
{h<, deniHllmi
»/(i™n)
amiof Jf =
= bXk.
(See Art. 1.)
Ex. 1. Find,
audS ft. wide.
Ex, 2. The ar
Fiud the biwe.
Ase of this theorem, the probiem of liuding tbe
reduced to tlie simpler problem of meaaurinj;
ons of the reclaiigle and taking their product.
i square feet, the
X of a reeUnglB is i
ingle 8 yds. long
,s altitude is 5 ft.
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PROTOblTION IV. TliEOHE.M
385. The area of a parallelofjmm if ''<juid to thi; product
of its base by ils aUitude.
Given fcho OJ ABCD with the baso AT) (douote.! by h)
and the altitude DF (denoted by It).
To prove area of ABCB = h X h.
Proof. From ^i draw AK || DF. and meeting CB pro-
duced, at K.
Tlien AK ± GK. Art. 133.
.■. AZFD is a rectangle witli base band altitude /(.(Why!)
In the rt. A AKB and 1)FG,
AB^BG.
AK=BF.
:. A AKB^A t)FG.
To each of these equals add the figure AliFD;
Then rectangle A7i'i^O oro ABVD.
But area of the rectangle AEFI)=b X h.
.-. nrcaa ABCD= h x /*■
Q.E.D.
386. Cor. 1. Parallelograms which have equal bases
and equal altitudes are eqmvalent.
387. Cob. 2. Parallelograms which. have equal bases
are to each other as their altitttdes;
Parallelograms ichich have equal altitudes are to each
other as their bases.
388. Cor. 3. Any two parallelograms are to each other
as the products of their bases a7id alliiadtin.
(WLy!)
(Why 7)
(WhyT)
Ax. 2.
Art. 383.
(Why ?)
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AEEAS OF POLYGONS 3d7
Proposition V. Theorem
889. The area of a triangle is equal to one-half the
product of its base by its altitude.
Given the A ARC with tlio base AC {denoted by i), and
the altitude Fli (aenoted by /().
To prove area of A ABC^i h X Ji.
Proof. Draw BD \\ AC, and CD ]] AB, forming the /Z7
ABDC.
Then BO is a diagonal of HJ ABJ)C.
:. A ABC^hCJ ABDC. Art. ir.fi.
But areaC7A£DC=6 X ft. (Why?)
.-. area A ABC=i 6 X ft. Ax. 5.
Q. £. D.
890. Cob. 1. Triangles which have equal bases and
equal altitudes {or which have equal bases and their vertices
in a line parallel to the base) are equivalent.
391. Cor. 2. Triangles n-htch have equal bases are to .
fach other as their altitudes;
Triangles which have equal altitudes are to each other as
their bases.
392. Cor. 3. Any two triangles an- to each other as the
prodticts of their bases and altitudes.
Ex. 1. Find the area ot a parallelogram whosa liuaa \s 'J tt. 8 ir..
and whose altitude is 2 ft. 3 in.
Ex. 2. Piud th« altitudu of a trian^'lB wIiu^b iiica is 180 sq. hi.
ana whose base is 1 ft. 3 in.
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2u8 BOOK IV. WANE (iC(IMi:'r]:V
PRoi'osiTioN VI. Theorem
393. If a, h, c denote the sides of a tnuiigle opposite the
angles A, B, C, respectively, and s = i {a + h -{- c) , the area
of the triangle = V^s (s — a) (s — b) (s — c).
Given the A ABC -withthe sides opposite A A, B and C,
denoted by a, b and c, respectively, i (a + b -\- c)
denoted by s, iiiid A an acute angle.
To prove area A AB(7 = V^s {s — .i) (s-h) (.--^T
Pioof. Draw the altitude BB and denote BB by h.
Then a-^b- + c- — 2h X AB. Art. 349.
.-. 2&X AB^b' + c^~<i:'. Axs. 2,
....«.-±|^,
But h'-<'' — A-D'-=(c + Al>) U — AD) Art. 3<
=1"+^ — )\' — 2F^; •''■
. (i. + e + (0 (t + c — a) (»+!.-(!) (g — J + c)
, a+ 6 — c = 2s — 2c, etc.
Hyp,, Ass, 4, ;
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AREAS OF POLYOONy '2?i^
4 6- 4 6-
2\/,{s-a)U-b)U-c)
B\it area A ABC^h b X h. Art- :m.
•■. area AAiiC-v'i. (>,■ — «) (s — fi) {s~c). Ax, a.
q. £. T>.
Proposition VII. Theorem
394. The area of a frapczotd is ei^nal In
tm uf Hn hasen muUiplied by its altitude.
Given the trapezoid ABCB with the bases AT) and BO
(denoted by& and b'), and the altitude FB (denoted by h).
To prove area of AB<7D=i {h + h') X h.
Proof. Draw the diagonal HI).
Then area of A AIW= h b X it. (Why t)
And area of A BVD^i V X h. (Wliy?)
Adding, area of AHCl) = h (W -^ b) h. (Why?)
Q. E. D.
395. Cor. The urea of a trapezoid equals the product of
the ■iiD'd'dii of the trapezoid by the allitiide. For the njediaii of
a trajjuKoid equals one-half the sum of the bases (Art. 179).
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240
,ANE GKO^CETKY
S96. ycHOLiUM. The area of a polygon of four or more
sides can usually be found in one of several ways; as, by
dividing the polygon into triangles and lahhtg the sum of the.
aiv.as of tkf. triangles; or, by draiving the loiige^tl diagonal
of the polygon and drawing
perpendiculars to this diago-
•ual from the vertices which it
does not meet, and obtaining .
the smn of tlie- areas of the
triangles and frupfzoids thus
formed.
e 5, 6, and 7 in.
are IS and 10 iii..
Ex. 4. If tte area of a trapazoii] is 135, and i
18, find its altitude.
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COMPAKISON OP I'OLVGONS
comparisom of polygons
Proposition VIII. Theorem
897. If two trianffles have an angle of one equal to an
angle of the other, their areas are to each other as the pro-
ducts of the sides includhuj the equal angles.
Given the
To prove
, ABO and AJ)^' having Zi
AABC__ABXAC
in eommon.
A ABF ADXAF
Proof. Draw the line DC.
Then the A ABC and ADC may be regarded &s having
their bases in the line AU, and as having the uoinmon
vertex C.
A ABC A P.
■-■AAIW^AJy ^''■'''-
(Why t)
l tlie corresponding members of these equali-
A ABC ^ ABXAG
A ABF ADX AF
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IIOOK IV. PLANE GEOMETKY
1'EOPUSITION IX. TlIEOHEM
398. The arms of Iwu mnUar irkiitykn ar.
as the squares of unij two hwiiuloijous shies.
Given the similiii- & ABC and A'B'C with AB and A'B'
homologous sides.
A .4 lic _ ~n?
To prove — -~", — r=zz:-
^ AA'iJ'C" A'B'-
Proof. Draw the homologous altitudes CF and C'F' .
A ABC AB XCF AB ^ CF
Then
A A'B'O A'B'XC'F A'B' C'F' ' "
arc to each other as the pymhicts of Ihfir hmes and aliitutlei)).
CF A B
Substituting TTTT/ foi' its equal 7^77^'
A ABC ^ ^^ y^ --^J^
A A'B'V A'B' A'B'
AB-
~VB'-
Ex. ]. If a pair ot lioroologous siJes of two similar triangles
4 ft. aud 5 ft., find the ratio of the areas of the triangles,
Ex, 2. Prove Prop. IS Ijj use of Prop. YILI.
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compjIeison of polygons 2i6
Proposition X. Theohem
399. The areas of tivo similar polygons are to each other
s the squares of any two homologous sides.
Given tiie similar polygons ABCDE and A'B'C'D'E',
■with their areas denoted by S and S', respectively, and with
AB and A'B' any pair of homologons sides.
To prove S r S'=Xb^ : A'B''.
Proof. Draw the diagonals AG, AD and A'C, A'B' from
the homologons vertices A and A'.
These diagonals will divide the polygons into similar & .
Art. 32!l.
A ABC AB^
. ^ Art. 308.
_^A'B'G' A'B'"'
" A^'B't" V7^^7 AA'Ci)' Vl^V
AA'B'E'
(Why?)
A ABC ^ A AGB A ABB _
■'■ A A'B'V £\A'GB' AA'B'B' (^'^'y'^
A ABC + A AC I) + A ABB __ A AUG _ ^^^ .^^„
' A A'B'C'+ A A'G'B'-i- A A'B'E' A A'B'G'
. '"^ ^ A ABC ^^ ,,
•• «' A A'B'C"
. ^ = J^. ^, 1
^' -^I'-t;'' o. E. D.
Ei. If a pair of honiologouH sides o! two site
"at! 2 ft. , Cud the ratio of the areae o£ the polygoi
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I'LAXE C.EOMKTKV
Peoposition XI. Theorem
400. Til a right triangle, ilie square on the htjpotenu!
is eqtdvaknl to tht sum of the squares on the two legn.
Given AT> the square on AC the hypotenuse of the rt.
A ABC, and BF and BK the squares on the leys Ba
;md BC respectively.
To prove ADo BF + BK.
Proof. Thi'oiigh B draw BL \\ AB, and meeting KB in
B. Draw BE and FC.
Then i ABC and ABG are rt. A .
:. GBCls a straight line.
Tn the A BAE and FAC, AB=^AF, i
Also
But
(for AL hai
Also
ZBAB= IF AC,
(for cach^ZBAC + arl. .
:. A BAE=/\FAC.
d AB=Aa.
I. Why ?)
In like
Adding,
square BF^2 A FAC.
rectangle Ai— square BF.
rectangle Ci~sqaare BK.
AL + CL, or AB^BF+BK.
As. :
J A BAE).
(Why ?)
401. Cor. 2'ke square on either leg of a right triangle
is eqiiivalent to the nquare on the hypotenuse ditninished by
the square on the other leg.
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CONSTEUCTION I'iiOBLEMB
COKSTKUCTIOK PROBLEMS
Proposition XII. Problem
402. To construe/, a square equivalent to the sum of iwo
given squnrex.
c
A Ir
Given two squares S and S'.
To construct a square oquiviilciit to S 4- •'^'-
Construction. Constrnct a right angle BAC. ah. 574.
On one side of this angle -take AB equal to a side of S,
and on the other side take AG equal to a side of S'.
Draw BC.
On a line equal to BC construct the square B.
Then B is the square required.
Proof. B=BC^'^'Jb'^ + A(f. Arf.4r)0.
.-. B=^S+ iS" Ax. 8.
Q. E. F.
403. Cor. To construct a sqtiare e<inivule}i.l to the sum
of three or more ghm squares. At C in the above figure
erect a liue GJ) X BC (Art. 274), and equal to a side of
the thii-d given square. Draw DB. DB will be a side of
a square equivalent to the sum of three given squai'cs, etc.
Ez- 1. CoBstruot a square equivalent to ths s
■whose sides are i in. and 1 in., respectively.
Ex. 2. By use of Art. 40J, taking a given Um
v'S; also v/'l. For example, eonstraet
1 o( t
) squares
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^'■iO liOOK lY. PLANE GKOMETET
PiiOmsinoN Xin. 3'roblkm
404. Tomm'InicI a ^qiuirc equivalent (o the difference
of two yiven aquare^i.
B
B
Given the squares W aiul S'.
To construct n sqnarij equivi
and S'.
Construction. Coustn
hnit t.o iln; difTnrcncR of ;
■t'Ait&ngh BAR'.
Art. 274,
Ou one siilo J B take AB equal to a side of the smaller
given square 8'.
From B as a center with a radius BC, etjnal to a side of
the larger square, describe an are intersecting AK in G.
On a line eqnal to AC eoustruct the square R .
Then B is the square required.
Proof.
B = AC-^BC- — AB',
Art, 401.
{the nqiiare on either leg of a right triangle is equitiateni to the square on
the hypotenuse diminished by the square on the other leg).
Ex. Construct fl square e
whose sidtB are 1 in. and i ii
Q. E. F,
ilitTereuce of two aquarea
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CONSTRUCTION PROBLEMS ^H
Proposition XIV. Problem
405. To cimstriid a square eqxdmhnt to a given paml-
lelogram.
i) k -p
Given tlie CJ AliCT) with base S and altitude h.
To construct a sciuave equivalent to ABCD.
CoEstructioa, On the lino EO take EF (sqnal to h and
FG equal to 6.
On EG iis a diameter construct a semicircle. Art. 27a, Post. 3.
At F erect a X meeting the semieircumference at K.
Art. 274.
On a line equal to FK construct the square S.
Then S is the required square.
Proof. S^KF'.
But KF''^bXh, Art. S43.
(n ± from a«ij poiii! hi a cinyiimfcrnnw lo a i^iaJiii^lci- is n mean pyopor-
tioiial iftwecn lite sajotenls of the diameter).
But area CJ ABGD^b X h. (Why t)
.-. 6'oarea £Z7 ABGD.
Q. E. F.
406. Cor. To construct a square equivalent to a giien
triangle, eonstriiet the mean proportional between the base
, and half the altitude of the triangle and construct a square
oa ihis mean proportional.
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i8 BOOK IV. PLANE GEOMETRY
Peoposition XV. Problem
407. To construct a triangle cqmvaleid to a give
Given the polygon ABCBE.
To construct a triangle equivalent to ABCDE.
Construction. Let --i, B, (7 be iiny three consecutive ver-
tices in the given polygon.
Draw the diagonal AC.
Draw BF \\ AC (Art. 279), and meeting AE produced
at#.
Draw FC.
In the polygon FGDE take the three consecutive ver-
tiees C, B, E, and draw the diagonal CE.
Draw BG || CE, and meeting AE produced at G, Draw
CG.
Then A FCG is the triangle required.
Proof. A ABC->AAFG, Art. wo.
(having the same base AC, and their veriiees in a live BF |] the base) .
Also A AGE = A ACE. Went.
And A FGB = A ECG, Art. 390.
[hamng the same base CEand their vertices in line DG [j >Mse).
Adding, A ABC + A AGE + A ECB o A AFC +
A.1CB+ A ECO. As- 2.
Or polygon ABCDE<^ A FCG.
»}. ffi. F.
408. Cor. To construct a square equivnUnt to a gi^en
polygon, use Arts. 407 and 406.
Ei. Construct a triangle equivalent to a given hesagon.
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CONSTRUCTIOK PROBLEMS M\)
Proposition XVI. Pkoblem
409. To construct a rectangle equivalent to a given square,
and having the sum of its iase and altitude equal to a given
line.
r D.
Given the sf|uiire S and the line ,4 B.
To construct a rectangle equivalent to S, nnd having the
sum of its base and altitude equal to AB.
Construction, On AB as a diameter describe the semi-
circumference ADB. Art, 275, I'OBt. 3.
At the point A erect a X , AG, equal to a side of S.
Art. 274.
Through C draw a line || AB (Art. 279), and meeting
the cLryumference at I).
DrawDE J. AB, Art, 273.
Construct the rectangle R with a base equal to EB and
an altitude equal to AE.
Then B is the rectangle required.
Proof.
DB- = AJ'JXEB.
(Why f)
But
DE=CA.
(Wby ?)
.: CA^=^AEXEB.
(Why?)
Or
S=>B.
410. Cor. The above problem is eqiiivaieiit to the
problem: Given the sum and product of iivo lines, to con-
struct the lines.
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50 BOOK IV. I'LANE GIilOMETKY
Proposition XVII. PiioJiLEJi
411. To eon&iruct a rectangle equivalent to a given square,
'tid having ike difference of Us base and altitude equal to a
liceu line.
[ZZZl
Given the square S and the line Ali.
To construct a
difference of its 1:
ictangle equivalent to .S', and
ie and altitude equ<il to AB.
Construction. On AB as a diameter describe the c
fereuce ABBF. Art. 275, Post. 3.
At A erect the X AC equal to a side of S. Art. 274.
Draw C'F through the center 0, and meeting the circum-
ference at the points D and F.
Construct a rectangle E with base equal to CF and alti-
tude equal to CD.
Tht:ii li is the rectangle required.
Proof.
CF: CA = CA : CB,
Art. -isa.
.: CA'=CFX CIK
(Why !)
:. S^B.
(Why ?)
Also the difference of the base and altitude of B=
CF—CB = DF=AB.
Q. E. F. .
412. Cor. The above problem is equivalent to tbo
problem: Given ike difference and the product of two lines.
to constrtKt the Ums,
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CONSTRUCTION PROBLEMS ^Ol
Proposition XVIII. Problem
413. To construct a polygon similar to two given simi-
lar polygons, and equivalent to their sum.
B' K"""L
Given the similar polygons P and P'.
To construct a polygon similar to P and P', and equiva-
lent to their sum.
Construction. Take any two homologous sides, AB and
A!B', of P and P'.
Draw J/A'X KL (Art. 274). making l/i:=J.B, and ^i
= A'B'.
Draw ML.
On A"B", equal to ML, as a side homologous to AB
construct the polygon P" similar to P. An, 3T2.
Then P" is the polygon required.
Proof.
I" A«B«'' "'■" P" ^"B"''
Art. 399.
(a. „r,a,
o/e
'1-rt similar polygons ore k> mch other a.
their homologous siiles).
, P+F' lYf + A^'^
? Ihe
*9"a'"i^» 0/
Atldiii?.
As. a.
^P" ^ .1"P"^
But
jU"2i:^"+ A'i^ = i¥//,
Art, 346.
Or
AB--\- A'B'- 'a"B"^_i
As. 8.
As. 8.
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BOOK IV. I'LASE GEOMETKY
Proposition XIX. Proulem
414. To construct a square which shall have a given
ratio (o a given square.
B
A'-^:[yW"";c
Given the square S and the lines m a.nd n.
To construct a square which shall be to S In tlie ratio
Coastruction. Take AB equal to a side of S and draw
AF, making a eonveuieut angle with AB.
On AF take AD equal to m, and J>F equal to w.
Draw DB. Draw F€ \\ DB, meeting AB produced in C.
Art. 270.
On AC, as a diameter, construct a semicircumference
AKC. Art, 27-1, Post. 3.
At B erect si L BK meeting the semieireumference at Tf.
Art. 274.
Construct a square B' having a side equal to BK, or x.
Then &' is the square required.
Proof. ^^ = a X 6. (Why?)
Also a : h = m -. n. (Why?)
□„„„„ «_«!_«!_"_'». ,,. . .
S'
ab b
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OOKSTRUOTION PRORLEMa 2:)d
Proposition XX . Pboelem
415. To construct a polygon similar to a given polyg&ii,
ind having a given ratio to it.
Given the polygon 7" and the lines m and n.
To construct a polygon /-" which shuU be similar to P,
and be to P in the ratio n : m.
Construction. Construct a square which shu.!! be to clie
square on AB as n : ni. Art. m.
Let A'B' be a side of this square.
Upon A'B' as a side homologous to AB construct a
polygon P' similar to P. Art. 37'J.
Then 1" is the polygon required.
p_
AK
I'
375
AJP
-".
JTb'-
n
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PLANK GEOMETIIY
Pkoposition XXI. Problem
416. To construct a polygon similar to one given polygon ^
and iquiviiUnt to another given polygon.
O^
Givea the polygons P and Q.
To construct a polygon similar to P, and equivalent to Q.
Construction. Conatruet a square equivalent to P, and
let in tie one of its sides. Act. 408.
Construct a square equivalent to Q, and let n be one of
its sides. Art. 4oa
Construct >1'/J', the fourth proportional to ni, n, and AB
Art. 3GG
On vl'-B', as a side homologous to ^l/>', euiistruet a poly
gon P" similar to P. Art. a72,
Then P is the polygon required.
^ , P m^ Tli^ P
Proof. _= =.^^=-_.
Constr
, Arte. 314,399.
P P
As. 1.
.-. F^Q.
Art. 305.
Q. I. F.
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EXERCISES . THEOREMS 2 JO
EXERCISES. CROUP 4«>
THEOEF.>[S CONCERNING AREAS
Ei. 1. The dia{;oiial3 of a parallelogram divide the parallelogram'
into four equivalent tnangles,
Ex. 2. Any straight line drawn through the point o£ intersection
of the diagonals of u parallelogram divides the parallelogram into two
equivalent parts.
Ex, 3. If, in the triangle ABC, D aud F are the midpoints of the
sides AJi and AC, respeetively, the area of ADF equals one-fourth the
area of ABC.
[SuQ, Use Art. 397,]
Ex. 4. If the midpoints of two adja.'ent aides of a pruallologram
be joined, the area of the triangle so fonned equals oiii^-eighth the
area of the parallelogram.
Ex. 5. If, in the triangle ABC, I) and /■■ are the midpoints of the
sides AB and AC, respectively, the triangles ADC and AFB are equi-
valent,
Ex. 6. In a right triangle show, by obtaining expressions for the
area of the figure, that the product of the legs equals the product of
the hypotenuse by the altitude upon the hypotenuse.
Ex. 7. If two triangles are equivalent, and the altitude of one is
three times the altitude of the other, find the ratio of their bases.
Ex 8. If two isosceles triangles have their legs equal, and half
of the base of one equivalent to the altitude of the other, the tri-
angles are equivalent.
Ex. 9 If two triangles have an angle of one the supplement of
an angle of the other, their areas are to each other as the products of
the sides ineludinj; these angles.
(I b
Ex. 10. Prove geomiitrieally that (ii+ii) =" •!■'■ [ ~~1~
+2(ib.
Ex, 11. Similarly prove («—i)'=a' + i''~2u6.
Ex 12. Similarly prove (o+f)}(n—;>)=(i=—i/'.
Ex. 13. The line joining the midpoints of the parallel aid^s of u
trapezoid divides the tvupezoid into two i^iiuivitliini piirt^..
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ii56
"BOOK IV. TLANr; GKOMETUV
Ex. 14. The linea joiuing tlio miiipolTit of otio diagonal ot i
quadrilateral to the verticea not joined by
the diagonal divide the quadrila,teral into B T O
two equivalent parts. / /
Ex. IB. Given QR and TS passins
through P, any point on the diagonal AC
Df a CJ , QR\\ AD, and TS ]] All, prove
QBTP<:^PRDS.
Ex. 16. Given OH^OD; proveAABC^
AADC. I-et the pupif also state this as a
USE Of ACXLLIARY LINES
Ex, 1. Given AJICI) a ZZ7 and P any
point inside ABDC; prove APJD + APBG
= £:.PAB+APCD.
Ex, 2, The area of a triangle is equal to
one halt the product o£ its perimeter by the radius of the inscribed
Ex. 3 If the estremitiea of one leg of a trapenoid bo joined to the
midpoint of the other leg, the middle one of the three triangles thus
formed is equivalent to half the trapezoid.
Ex. 4. Tho area«f a trapezoid is equal to the product of one leg by
thn pi'rpeudicular on that leg from the midpoint of the other leg.
Ex 5. If the midpoints of the aides of a quadri-
lateral be joined in order, the parailelogiam thus
formed is equivalent to one-half the quadrilatetal
Ex. 6. A quadrilateral is equivalent to i triangle
two ot whose aides are the diagonals of the q u id i: lateral, the angle
included by these sides being equal to one of the angles formed by
the intersection of the diagonals.
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EXERCISES. THEOREMS J,i>i
EXERCISES. CROUP 4S
THEOEEMS PROVED BY VARIOUS METHODS
Ex. 1. If through the midpoint of one leg of a trapeioid a line be
drawn parallel to tde other leg to meet one base and the other base
produced, the parallelogram so formed is equivalent to the trapezoid,
Ex. 2, If the midpoints of two sides of a triangle be joined to any
point in the base, tbo quadrilateral so formed is eqaivalent to halt the
triangle.
Ex. 3. If P is any point on AC the diagonal of a paralleloeram
ABCD, the triangles AFB and APD are equivalent.
Ex. 4. If the siili) of an equilateral triangle be denoted by a, tha
area of the triangle equals -~r~-
Ei. 5. Find the ratio of the areas of two equilateral triangles, i£
the altitude of one equals the side of the other.
Ex. 6. If perpendiculars be drawn from any point within an
equilateral triangle to the three sides, their sum is equal to the alti-
tude of the triangle.
Ex. 7. If -E is the intersection of the diagonals AC and SD of a
quadrilateral, and the triangle ADE is equivalent to the A £EC, then
the liues A IS and CD are parallel.
Ex, 8. If, in the quadrilateral AliCD, the triangles AUC and AIiC
are equivalent, the diagonal AC bisects the diagonal BD.
Ex. 9. If two triangles have two sides of one equal to two sides
of the other, and the included angles supplementary, the triangles are
e(iuivalent.
Ex. 10. II, in the parallelogram -4lJ?CD,
F is the midpoint of the side BC, and AF
intersects BD in K, the triangle liEF=-f^ the
parallelogram ABCb.
Ex. 11. Given PQ [[ AC, and PR \\ AB; ^
prove AQAB a mean proportional between AHQP and AFRC.
Ex. 12. P Ik any point in the side llC
of liie parallelogram A/!(-lr and HP pro-
duced meets .1 /( produced iu <l Show that
the triangles BPA and CPQ are equivalent.
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258 BOOK IV. PLANE GKOMRTIiY
EXERCISES. GROUP 43
PROliLEMS IN CONSTRUCTING AREAS
Ex. 1. Construct ii square having twii;*; Ihi; nrea i)£ n given square.
Ex. 2. Construct a square hiiviug three times tho area of a given,
square.
Ex. 3. Construct a square oquivaient to tbe sum of three pivett
squares.
Ex. 4. Transform a given triangle info an equivalent isosceles
triangle having the same base.
Ex. 5. Transform a given triangle into an equivalent triangle
having the same base, but having a given angle atjjaeeut to the base.
Ex. 6, Transform a triangle into an equivalent triangle with the
same base, but naving another given side.
Ex. 7. Transform a parallelogram into an equivalent parallelo-
gram having the same base, but ooutaining a given angle.
Ex. S, Construct a triangle similar to a given triangle ami con-
taining twice the area.
To construct a similar triangle containing five times the area: how
is the construction changed !
Ex 9. Bisect a given triangle by a line parallel to the b^se.
Ex. 10. Constmet a polygon similar to tivo given i^imilar pohgons,
and equivalent to their difierence.
Ex.11. Draw a line parallel to one
side of a given rectangle, and
cutting off five-sevenths of the area.
Ex.12. Bisect a parallelogram by a
lino perpendicular to the base.
Ex. 13. Through any given point c
km , line U„Mme , t-iyea
parallelogram.
Ex. a. Construct a triangle equivalent to
a given triangle, ABC, having a given base
AD, but the ZBAG adjacent to the base un-
changed.
[SuQ, Draw CE \\ DU, etc.]
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EXEHCISES. PEOBLEMS 259
B:i. 15. Transtorm a parftllelogram into an equivalent parallelo-
giaxa haviagagivenbaee, but the angle adjacent to the base unchanged.
Ex. 16. Tranaforra a given triangle into an equivalent riglit
triangle having a given leg.
[SUO, Uhb Ex. 14, then Ex. 5,]
Ex, 17. Transform a given triangle into an equivalent right tri-
angle having a given hypotenuse.
[SuG. Piod the altitude upon tlie hypotenuse of the new triangle
by fijsding the fourth propovtionnl to what three liiu's ? ]
Ex. 18. Transform are-entrant pentagon into ;
equivalent triangle.
Ex. 19. TranBform a given tnaugle ioto
equivalent equilateral triangle.
[Sue. See Art. 416.]
Ex. 20. Biseat a triangle by a line perpendicular to one of its aides.
[Sue. See Art. 416,]
Ex. 21. Construct a square equivalfiit to two-thivila of a given
square .
EXERCISES. CROUP 44
PROBLEMS SOLVED BY AEUEBHAIC ANALYSIS
Ex. ]. Transform a given rectangle into an equivalent rectangle
with a given base.
Ex. 2. Transform a given square i;
given leg.
Ex. 3. TrauBform a given triangle
right triangle.
ato a right triangle having a
into an equivalent iaosneiea
Ex. i. Draw a line cutting off from
triangle equivalent to one-half the give
[SuG. Use Art. 3SIT.]
, a given triangle an isosceles
a triangle.
Ex. 5. Through a given point in
draw a line biaecting the area of the tri
one side of a given tz'iangle
angle.
Ex. 6. Transform a giveu square into a rectangle which shall have
three timeu th« jietimetec oi the given square.
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Zbi) BOOK IV. PLANE GF.OMETHY
EXERCISES. CROUP 4S
PR0RLE3IS S0LVI3D BY VARIOrS >[ETHODS
Ex, 1, Bisect a given, paralle log vara by a line parallel to the base.
El. 2. Tmnsforni a pavallBlogr.am into an equivalent parallelo-
gram having the aame base and a given side adjacent to the base.
Ex. 3. Construct a square which shall aontain four-seventhg of
the area of a given square.
Ex.4.
In two different ways
i constr
uct a square t
.aving
three
times the
areaof aglvensfjuare.
Ex.6.
Trisect a given triangl
e by line
IS piiraliel to thi
i base.
Ex. 6.
Find a point ivithin a
triangle
such that line;
i drawn
from
it to the \
'erticestriseet the area.
Ex. 7.
Find a point within a
triangle
such that iinci
•. drawn
from
it to tha
three vertices divide the area i
iito parts whic
■h shall have
the ratio 1
2:3:4.
[Sue.
II one of the small A
eontains
i tlie area o£ .
original
A, a
line thcoi
igh its vertex cuts off s
the altitude, etc.]
Ex. 8.
Divide a triangle ini
to tbree
equivalent parts by
lines
Ex. 9. Divide a triangle into three equivalent parts by lines drawn
through a given point J' in one o£ the sides.
[Suo. Use Art. 307.]
Ex. 10. Divide a giveu qiiadrilatei-al into three equivalent parts
by lines drawn through a given vertes.
Ex. 11. Through a given point in tlie G
base of atrapezoid draw a linebiseeting the ,' ''.,
area of the trapezoid. / \
Ex. 12. Bisect the area of a trapezoid y ' \ p
by a line drawn parallel to the bases, p[ \ p
[SUG. Construct A GKF similar to A J \ p
ABG aud equivalent to i num el wliat two
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Book V
REGULAE POLYGONS. MEASUREMENT OF
THE CIRCLE
417, Def, a regular polygon is a polygon tliat U both
equihiteral aod equiangular.
Proposition 1 . Theoreie
418. An equilateral poiyijnii (hat in in'^i-rihed in o circle
is ulso eqaiaiigtilar and regular.
Given ABC . . .K an inserilied polj^goii, with its sides
at;, nc, CD, etc., etiiia!.
To prove the polygon ABC...K equiangular and
rogular.
Proof. Arc A7.' = ai-c mj^avc- Cl>, ptc. Art. 21M.
.-. are ABG^ayq. BCI}=arc CJ>E, etc. Ax. 2.
.-. ^ ABC^IBCI)^ ^CDE, etc.. Art. 2C0.
(all 1 inscribed in the same segment^ or in equal segments, are equal).
.: tlie polygon ABC , . . Jt is equiangiiki'.
/. the polygon ABC . . . KU regular, Art, 417.
0, z. B.
{■JCI)
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262
KOOK V. PLANE GEOMF.TKY
419. Cor. 1. // the arcs suhtended by the sides of a
regular insrrih''(l polygon he bisected, and each point of
bisection lie joined to the nearest vertices of the polygon, a
regular inscribed polygon of double the number of sides is
formed.
420. Cob. 3. The perimeter of an inscribed polygon is
less than the perimeter of an inscribed polygon of double ike
number of sides.
Proposition II. Tiif-orem
421. A circle may be circumscribed about, and a circle
may be inscribed in, any regular polygon.
Given the regular polygon ABCIIE.
To prove that a O may be circi-imBeribeil about, or in-
scribed in, ABCBE.
Proof. I. Throueh A, B aad C (Fig. 1) , any three suc-
cessive vertices of the polygon ABCDE, pass a circum-
ference. Art. 235.
Let be the center of this eircumferenee.
Draw the radii OA, OB, 00. Also draw the line OD.
Then, in A OBG, OB=OG. (Whj-r)
:. lOBC^lOCB. (Whjf).
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KEGULAE POLYGONS 263
But Z ABG = Z BCD, Art. 417.
(being A of a regular j)Olii3on).
Subtracting, Z OBA = Z 00i>. (Why?)
Hence, iu the A OAli and OCD,
OB=OG. (Whyt)
^B-Oi). (Why?)
Z OBA = Z OC'D,
(JH;;! jtniuecf).
.-. A ABO^AOVB. (Why!)
.-. 01>=0A. (Why?)
Hence the cireiirafereuce whieli passes through the
vertices A, B and C, will iilso pass throuirh the viirtes I).
In like manner, it may bo proved that this circumference
wdl pass tbrongh the vertex E.
Hence a circle described with as a center, and OA as
a radius, will be circumscribed about the given polygon.
H. The sides of the polygon ABODE (Fig. 2) are equal
chords iu the circle 0.
Hence they are equidistant from the center. Art. 22(i.
.■. a circle described with as a center, and the distance
from to one of the aides of the polygon as a radius, will
he inscribed in the given polygon.
q. E. B.
422. Def. The center of a regular polygon is the com-
mon center of the inscribed and circumscribed circles, as
the point in the above figure,
423. Def, The radius of a regular polygon is the radius
of the circumscribed circle, aa OA in the above figure.
424. T>FF. The apothem of a regular polygon is the
radius uf tliL- inscribed circle.
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2fJ4 BOOK V. PLANE GEOMETRY
425. Def. The angle at the center of a regular polygon
is the angle between two radii drawn to tlie extremities of
any side, as the augle AOB.
426. COE. The angle id the. center of a re</Hlar polygon
is equal to four right angles divided l»j the ntiinher of sides.
Hence, if n denote the number of sides iu the polygon,
the angle at the center of a regular jiolygon eijuulx —~ — ;
also Ike angle between un apotkem and the nearest radius,
in a regular polyjoit of ii sidvs, eqiiaU
PHO^osITIO^f III. Theorem
427. If the circumference of a circle be dirided info any
number of equal ares,
I. The chords of these arcs form a regular inscribed
polygon;
II. Tangents to the circrimference at the points of di>:ision
form a regtdar circumscribed polygon .
e^:^c
Given the circumference ABC, divided into the equal
arcs AB, BC, GJ), etc., the chorda AB, BC. etc., and PQ,
QB, etc., lines tangent to the circle at B, (J, etc.
To prove ABODE a regular insei-ibcd polygon, and
PQRST a regular circumscribed polygon.
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eegUlak polygons Libu
Proof. I. The chords AB, BC, CD, etc., are equal.
(Why r)
.■- po'ygoii ABODE is equilateral and regular. Art. 418.
II. Ill the A APH, BQG, CRD, etc.,
AB=liC=Cn, etc. (-Why!)
Also ZP.4B- ZPfiA = IQIiC^ IQGB= ^ BCD, etc.
Art. 264.
{«id< la:.
.'. A AZ-'7i, /iX'<-''. Clilt, lite,., areeqiiiil, isosceles triangles.
(Why?)
.-. ^1'-^ IQ ^ Z7i*, etc. (Why?)
Afld AP^PB^BQ=QC, etc. (WhyT)
.-. PQ=QB = nS, etc. Ax. 4.
.-. PQBST is a regular i)Cilygoa. Art. 417.
Q. E. ».
428. Cor. 1. If the arcs AB, BC, CB, etc., be bisected,
and (1 tangent fie draimi at each point of bisection , a circum-
scribed regular polygon of double the number of sides of
PQI18T will be formed.
429. Cor. 2. The perimeter of a circumscribed regvlar
polygon is greater than that of a circumscribed regular poly-
gon of double the number of sides.
Ex, 1. Fiud the number o!^ <Jep:vees lu the central angle of a
legalar poiitagon. Of a regular hexagou. Of a square.
Ex. 2. WUat is the sliovt name for an iuscribeii equilateral quad-
Tilu^al ;
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26G BOOK V. PLAVE GEOMETRY
Proposition IV. Theorem
430. Tangents to a circle <at the midpabits of the arc?
subtended hij the sides of a regular inscrihed pobj'jon form a
regular circumscribed polygon whose sides U7'€ purulUl to
the corresponding sides of the inscribed polygon.
Given the regular polygon ABODE iiiscribed in tlie O
ACD; P, Q, B, etc., tiie midpoints o£ tiie arcs AB, BO,
CJ), etc.; and A'B', B'G', CD', etc., tangents to the circle
at P, g, E, etc.
To prove A'B'C'D'E' a regular polygon with its sides ||
corresponding sides of the polygon ABCDB.
Prool The ares AB, BC, CD, etc., are equal. Ait. 218,
.-. the arcs AF, FB, BQ, QG, etc., are equal. Ax. 5.
.■, the arcs I'Q, Qli, BS, etc., are equal. Ax. 4.
.■. A'B'C'D'E' is a regular polygon, Art. 427.
{if Ike cii-ciimfereiiee of a O ho diiikkd, clc.).
Side AB A OP. Art. us.
A'B' ± X>P. (Whyt)
.-. AB II A'B'. (Why?)
In like manner, each pair of homologous sides in the
two polvgons is parallel.
Q. E. D.
431. Cob. Homologous radii of an inscribed and a eir-
eumscribed regular polygon, wlhse sides are parallel, coin-
cide in direction. Thus, in the above figure A POA and
';. 42e°
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EEGnLAB POLYGONS
Proposition V. Theorem
432. Two regular polygons of the same number of sides
re similar.
Given K and K' two regular poJyf>ons, eaoli of n sides.
To prove if and K' similar.
Proof. Eaoh Z of A'^ -■■ '' '" " -■-- -" ■■- • - -. Art, 174
(i» an equiangular pohjgon ff n sides, each Z = — - — - — )
Similarly each Z of IC— " " ' —
Hence K and K' are mutually equiaog'ular. As, i, Art. igs
AB_
Also
AB = BC .-. fi;,.l. ,
Art. 417, Ax, ii.
And
a-b:b'<7 ,. ff;.l.
(Why!)
. AB^A'B' ^|, AB _ BO
"" fiC B'C' A'B' B'C'
(Why ?)
In like
BO CD BE ^
manner ^gJ^^-p^-J^,, elo.
Henee K and K' have tlieir homologous
sides propor-
Hence K and K' are similar.
AH. 221.
Q. E. D.
433. Cor. The areas of lico regular polygons of the
same mmibe.r of sides are to each other as the squares of any
two homoloyoHs sides.
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2G8 BOOK V. TLANE GEOMETRY
Pkopositiox VI. Theokeji
434. I. The perimeters of ttco regular polygons of the
same numher of sides are to each other as the radii of their
circumscribed circles, or as the radii of their inscribed
circles;
11. Their areas are to each other as the squares of these
radii.
Given AC and A'C two regular polygous, each of n
sides, with centers and C, and with perimeters denoted
by P and J", radii by E and R', and apothems by r and »■',
respectively.
To prove. I. P : F' = R -. E' = r -. r'.
II. Area AO -. area A'C' = R^ : R'-^r- -. r'-.
Proof. I. Thepolygons ACand A'Care similar. Art. 432.
Hence P : P-^AB -. A'B'. Art. 34T.
But, in the A OAB and (yA'B',
I AOB = lA'O'B', [foreai-h Z=l^). Art, 42G.
Also OA : OB^O'A' : O'B', (for each A is isosceles).
.-. A OAB and O'A'B' are similar. Art. 327.
.-. AB : A'B'=OA : CfA'. Art. 321.
And AB : A'B'=OL i O'L'. Art. 338.
.-. P : F^OA : O'A'^OL : O'L'. Ak. i.
Or F: P'^B: R'^r: r'.
II. Area AC : area A'C'='AB^ -. A'B'^. Art. 399.
But AB^ ■.A^'- = R' : R'" = r^ : r'-. Art. 314.
.'. area AG ; area A'C'^B- : £'• ; r : r'^. As, l.
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EEGULAE POLyGONS ZW
Proposition VII. Theorem
435. If the numher of sides of a regular inscribed poly-
gon he indefinitely increased, the apothem of the polygon
approaches the radius as a limit.
Given the regular inscribed polygon AB . . . D oi n sides,
with radius OA and apothem OL.
To prove that, as n is indeiinituly Ln(;reascd, OL ap-
proaches OA as a limit.
Proof. In the A OAL, AL> OA — OL, Art. 93.
{any siite of a ^ is greater than the lUffcrencn hchrccii the
other two sides.)
But, asn^^,AB = .: AL^O. Art, 25a, 3.
Hence OA—OL = 0.
:. OL = OA, or )- = JJ.
Oi' the limit of thf apotliera OL is the nidius OA.
Q. s. J>.
436. Cor. As« = ^, iJ^— f- = 0.
For, iS^ — »-2={K + r) (iJ — r). But, as »(=»,
B+r^R + E or 2 R, and [i — r = 0. As, s.
.-. S2^,-- = 2i," X 01- 0. Ai<. s.
Ex. A pjiir of bomologous sidfis of two regiiiai' pentagans are 2
and 3 ft. Find ttn) ratio of tlie areaa of llio yglygons.
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liOOK V. PLANE GEOMETRY
Proposition VIII. Theorem
437. The hnffHi of any line incloning Ihe circrim/crence
of II circle, and Jtot passing within the circumference, is greater
than tlie length of the circumference.
Given the cirenmferenee AKEF, and ABCDEF any line
which does not pass within AKEF.
To prove cireumferenee AKEF < perimeter ABCDEF.
Proof. Let ST be any point on the circumference AKEF
not touched by the line ACBEF.
At K draw a tangent to the given circle, meeting ACEE
at B and I>.
Then the straight line BKB < line BCD. Art. 15.
To each of these unequals add the line IlEFAB.
Then line ABKDEF < line ACBEF. Ax. 9.
Henee every, line enveloping the circular area AKEF,
except the circumference AKEF, may be shortened.
Hence the circamference AKEF is shorter than any line
enveloping it.
'^ " Q, S. B.,
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REGULAR POLYGONS 271
438. Cor. 1. The circumfef&iice of a circle is less than
the perimeter of any polygon circumscribed about the circle.
4.89. Cor, 2. The circumference of a circle is greater
than the perimeter of any polygon inscribed in the circle,
for each side of an, inscribed polygou is less than the aro
subtended by it.
440. Cor. 3, The difference between (he perimeters of
an inscribed and a circumscribed polygon is greater than the
difference between either perimeter and the circumference of
the circle.
Proposition IX. Theorem
441. If the number of sides of a regular inscribed, or
of a regular circumscribed polygon be indefinitely increased,
I. The perimeter of each polygon approaches the circum-
ference as a limit;
II. The area of each polygon approaches the area of the
circle as a limit.
Given a circle of eircnmfei-ence (7 and area A, with regu-
lar inscribed and cireumscribed polygons, each of n sides,
with their periraetei's denoted by P and P', and tijeir areas
by AT and K', respectively.
To prove that as n is indefinitely increased, P and P'
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2i2 BOOK V. PLAXE c.T-,o^iEn;Y
each approaches C as a limit,, au,l K and K' each ap-
proaches A as a limit.
Proof I. Denote the apothems of tlie two polygons by
H and )■.
Then j,=''.- Art-™-
Hence ^^ =^^- '^''- ^^"■
Let 11 = 0=; then Ji—r = 0. Art. 43ii.
Rut p'— < f"— 7' (Art. 440) .-. 1"—C^0, or F = C.
Also C— P < i"— .?' (Art, 440) .-. C — P=n, or P^C.
K
Let H =1. :o ; then 7^- - (■= i 0. An. 43G.
... ^^'" "T . ..ll i 0. Art- 253, 4.
(/(J)' Wic d(noinin'i(or , J'', infciscs).
H^nee — J=-^ i (As S), Z. K' ~ K= 0, (Art. 253, i).
But ,ff'— ^1 < iL'— .ff(Ax. 7) .-. .B:'— A=0, or ff' = J..
Al::;o A — K<K'—K{k:^.r) :. A — K=0, or K^A.
Q. E. D.
Ex, Find the perimeter and firea of a square field, one of whose
BldBS is 10 rods. Find the same in a square iield, oni: of wlioso sides
is 20 rids. Is it more economical, thei'eforf, to leuee land in largs
oc small gelds t
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REGULAR POLYGONS
Proposition X. Theorem
442. Two circumferences have the same ratio as (kei
radii, or as thHr dia
Given the circles and (7, with eireuniferenees denoted
by Caiid C , mdii by R and }}', diameters by D and i/, re-
Kpoci.ively.
To prove C : C = R : R' = D : D'.
Proof. Ill the given ® let regular polygons of the same
number of sides be inscribed. Denote the perimeters of the
inscribed polygons by P and P', respectively.
Then
F' It'
lience, by alternation, JT^jv,' ■A"''- 30'/.
If, now, the number of sides of the similar inscribed poly-
gons be increased indefinitely, P becomes a variable ap-
proaching C as a limit. Art. 441.
P
Hence— becomes a variable approaching — as a liri
' E
' Ji"
But the v;
riable
— =the variable
C
c
R
' R'
:. c : e-
-R: R
Also %-,--
VL
Art. 315.)
('
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2 1 4 ROOK y. I'LANK flKOMiyi'KY
rROPOSITlON XI. TnEOIlKM
443. In Cirri/ circle, the ratio of the circumfcrenre to Ikc^.
diameter (that is, the nuniber of times the diameter is
contained in the circumference) is the same, and may be
i by an appropriate symbol {ti).
Proof.
Given two © with cireumferences denoted by G and C,
and diametei's by V and D', respectively.
To prove | -|-»
C D
By alternation Jt^T/ Art- 30T.
Hence, in any given circle, — has a vahic equal to that
.■. the value of — is the same in all circles.
Denoting this constant by ti, in every eircle"^ = 7t.
Q. £. B.
444. Formula for the circumference in terms of the
ladius.
1)
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EEGULAU POLYGONS
445. Formula for length of arc of a circle.
arc _ central Z
circiiiufereEoe 360°
PiiorosiTioN XII. Thkorkh
446, TJir arra of a regular polyjon is equal to oiie-luilf
llhe product of its peritiieter hy its iipoihem.
Given the regnhir polygon ABGJ>E with area denoted
by K, perimeter by I', and apothem by r.
To prove £=JPX?-.
Proof. Draw the radii OA, OB, 00, etc., dividing tho
polygon into as many A os, the polygon has sides.
Ail the ^ thus formed have the same altitude, r. Art.308.
.■. the area of each A = i product of its base by r.
Art. 3S9.
Hence the sum of the areas of the & = | product of the
sum of the bases of the A by r, or= % PX-r.
But the sum of the areas of the & equals the area of
the polygon. Ax. 6,
Hence A'='. V X )■. An. s.
Q. E, D.
447. Def. Similar sectors are sectors iu different cir-
cles which have e{|ual angles at the center.
448. Def. Similar segments are segments in different
circles whose area subtend o'liud angles ut the center.
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TliOrOSITIOX XIII. TUEOKEM
449. T!ir arm nf a circle is equal to one-half the prod-
i€t of it-1 circumference hij ils radius.
Given a O with fiircunifciTiiee denoted Ity C, radius by
E, and avea h_v K.
To prove K= i C X B.
Proof. Cireiimscvibe a regular polygon about the given
eircle, and denote its perimeter by P and its area by K'.
In this ease the apothem of the regular polygon is R.
Hence E'^iPXK. Art. 44fi.
Let the number of sides of the circumscribed polygon
be increased indefinitely; then
iT' becomes a variable approaching £^ as its limit; Art. 441,
Pbeeomes avariable approaching Gas its limit; Art. 441.
And viiriiilik'^i'Xfliipproitches I Ox ^iis a limit. Art. 253,2.
K' = i i' X A' always. Art 446.
But
Uenc:
--\Cx R.
0. E. I
450. Formula for the area of a circle in terras of the
radius li.
K^i CXB; but C=27lR. Art. 444.
.-. -i(27r-fi)-K. Or K^TiE''. Ax. e.
Again E-J i> .-. .fi^^-^-- As. 8.
461. Cor, 1. The area of a circle is equal to the square
of the radius, multiplied hy n; or to one-fourth the sgiiare o/
the diameter, muUiplied hy %.
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MEASUEBMENT OF THE CIECLE 277
452. Cor, 2. The areas of two circles are to each other
as the squares of their radii, or us the squares of their diam'
eters.
K TtIP E- K inJ)'^ D"
For
453. The area of a sector is equal to one-half the prod-
uct of its radius by its arc.
1 { ceiiti-al I \
454. Cor. 3. Similar sedor^
squares of their radii.
Ex. 1. The measurement of tiie area of a circle can be reduced
to the measurement of the length of what single straight line ! Can
it he reduced to the measurement of any otiiev single straight line I
»nt of a sijigle curved line?
Find the area of a circle,
Ex. 2, Whose radius is 10 ft. Ex. 6. Whose radius is 2li ft.
Ex. 3. Whose diameter is 10 ft. Ex. 7. Wliose radius is E; IB;
Ex, 4. ■\Vhose radius is li ft. Et, 8. Wliose radius is Ei/5.
Ex. 5. Whose radius is iJ> ft. Ex.9. Whose diameter is *Bi/3.
Ex. 10. If the radius of oue cirele is 10 times as gfeat as the ra-
dius of another circle, how do their areas compared Also bjw do
their circumferouees compare ?
Ex. II. A wheel with 6 cogs is geared to awheel with 4S eogs,
How many rovolutious will the sniaUer wlieel make while the larger
Wheel revolves once ?
Ex. 12. A 2 in, pipe will diseharKe how much more water in a
^ven time thau a 1 in, liipij f
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2 /« BOOK V. PLANE GEOMETliY
Propositiom XIV. Theorem
455, 77»' areas of iito similar serjmeutsare tcreach other
as (he squares of their radii.
A
Given the siraiiar segments ABC and A'B'C in ciroici,
'liose radii are B and R'.
To prove segment ABC -. segment A'B'C' = B~ It'"^.
Proof. Draw the raGii OA, OB, O'A', O'B'.
Then the sectors OAB and (yA'B' are similar.
And A OAB and O'A'B' are similar.
. sector OAB R' , A 0.4 R
I. 448, 447.
Alii. 327.
t. 454, 338.
" scetoc O'A'B' E'-' A O'A'B'
. seetor OAB _ A OAB
" sector O'A'B' A O'A'ii''
. sector 0-4 B i sector O'A'B'
" A O^ii A O'A'B'
. sector OAB— A OAJJ__sector O'A'B'— A O'A'B'
segment ABO __ segment A'B'C
A OAB ~ A 0'.d'B'
;ment ABO _ f A OAB \ ^Kf .
meat A'B'C \A O'A'B' J W^
rts. 307,308.
Q. £. I).
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CONSTRUCTION PBOBLBMS
CONSTRUCTION PROBLEMS
Proposition XV. Problem
456. To inscrihe a square in a given circle.
Let the pupil supply a aolutiou.
457. Cor. By Msecting the arcs AC, CB, BD, etc.,
and drawing chords, a regular octagon may ie inscribed i»
the circle; by repeating the process, regular polygons 0/I6,
82, 64, . , . and 2" sides may be inscribed, where n is a
positive integer greater than 1.
How eau a regular polygon of 2" sides be cireuinscribed
aboiit a given circle 1
PROPOiilTION XVI. PROBLEJI
inscribe a regular JK^xagon in a given, circle.
Let tbe pupil aiipply a Hointion.
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BOOK V. ri.ASE GEOMETllY
of a regular innrrihed hexagon
459. Cor. 1. The si<
equals the radius of the c
460. Cor. 2. By joining ihe alternate vertices of a
regular inscribed hexagon, an equilateral triangle can he in-
scribed in a given circle.
461. Cor. 3. By bisecting the arcs AB, BC, CD, etc.,
and drawing chords, a regular polygon of 12 sides can be
inscribed in a given circle; by repeating the process, regular
polygons of 24, 48, ... 3 X 2" sides can be inscribed.
Similarly, how can a regular polygon of 3 X 2" sides
be circumscribed about a given circle ?
Proposition XVII. Problem
462. To inscribe a regular decagon in a given circle.
Given the circle 0.
To inscribe a regular decagon in the given O.
Construction. Draw any radius OA, and divide OA in
extreme aud mean ratio at K, OK being tbe greater seg-
ment. Art. 371.
With A as a center and OK as a radius, describe an arc
cutting the given circumference at B.
Then the chord AB J5 a side of the decagon required,
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CONSTRUCTION PROBLEMS 281
Proof. Draw OB and KB.
Then OA -. OE^OK : KA, Constr.
But AB^OK, .: OA : AB-=AB -. KA. Ax. y.
Also lOAB^lKAB. idont.
.-. A OAB aud EAB are similar. Art. 327.
/. ^0 — ^ABK [homolog. A of similar &).
And A AKB is isosceles, Art. 321.
[bdiig simHar lo A AOB, ttfticA is isosceles),
;. AB^KB=KO. Ax- 1.
Ileiicu 10= IKBO. (Why?)
J .-. A0AB=2 10. Art. yti.
[ 10^ 10.
Adding, Z AiJO + Z OAS + Z 0- 5 Z 0.
But Z ABO + Z OAB + Z 0=2 rt Z . Art. 134,
.-. 5 Z0=2rt. A. Ax. I.
.-. Z 0=i of 2 rt. A , or iV of 4 rt. A .
:, are AB is iV of the eiretimf ereneo ,
Hence, if the chord AB be applied ten times in suc-
cession to the circumference, a regular decagon will be
inscribed in the given circle. An, 418.
Q. E. F.
463. OoR. 1. By joining the alternate vetikes of a
regular inscribed decagon, a regular pentagon can be inscribed
in a given circle.
464. Cor. 2. By bisecting the arcs AB, BC . . . and
draiving chords, a reyular polygon of .20 sides can he in-
scribed in a given circle; by repeating iJte process a regular
polygon of iO, 80, ... 5 X 2" sides can be inscribed.
How can a regular polygon of 5 X 2" sides be cireuia-
scribed about a given circle )
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28!^ liOOK V. PLASE OEOMKTHV
Pkoposition XVIII. Pkoblej!
465. To inscribe a regular polijgon of fifteen sides (pen-
tedecagon) in a given circle.
Given the O AK.
To inscribe a regular pentedecagon in the given O .
Construction. Draw the chord AC equal to the radios of
the given O, and the chord AB equal to a side of a regular
decagon inscribed in the circle. Art. 462.
Draw the chord BC.
Then BC is a side of the required pentedeeagon.
Proof. .AC is a side of a regiiiar inscribed hexagon.
Art. 459.
.■, arc AC=i of the circumference. ,
In like manner are AB = iVof the circumference. Art. 4G2.
.", arc BO—t — "i\, or tV of the circumference. As. 3.
Hence, if chord BC be applied fifteen times in succes-
sion to the circumference, a regular pentedeeagon will be
inscribed in the given circle. Art. 418,
Q. E. F.
466. Cor. By bisecting the arcs BO, CD, . . . eic,
drawing chords, and repeating the process, regular polygons
of '60, 60, . , . 15 X 2" sides can M inscribed in a given
circle.
How can a regular polygon of 15 X 2" sides be circum-
scribed about a given circle !
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COMPUTATION PROBLEMS 283
COMPUTATION PKOBLEHS
Proposition XIX. Problem
467. Given the side and raditis of a regular inscribed
polygon, to find the side of a regular inscribed polygon ofdouUe
the nuttier of sides, in terms of the given quantities.
Given the circle with radius R, AB a side of a regular
inscribed polygon, and AC a side of the regular inscribed
polygon of double the number of sides.
To determine AC in terms of AB and E.
Solution. Draw the radii OA and OC.
Then OC is the X bisector of AB. (Why ()
But arc ACB < semieircumferenee. (Why ?)
.: AC < a quadrant. Ax. 10.
.*. Zj4 00 is an acute Z. Art. 257.
Henee,in A 010, Zc^=OP+0C^— 2 OCX 07'.
Or AC^^2E^ — 2BXOD
But, in the rt. A OAD, 7)& ^OA^—AlP , Art. 347.
Or OF = R' — \iABY-. Ax. 8.
.: OD^V'E-~i JF^i V'ili^ — AB'.
Substituting for OD its value thus obtained,
A<f = 2E^ — R-l/i^ —A^.
AC-= VKi^E—ViB^ — Aff),
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284 BOOK V. I'LAXE GEOMETRY
468. Special Formulas. If tho radius B, lie takt>n as 1.
AC = \^2 — y4: — AB\
If a side of tlie regular inscribed polygon of n sides be
denoted by S„, and ^=1, then S-y,,^^'^ — VT^^Sl
Proposition XX. Problem
To compute approximately the numerical value ofn.
Given a O whose radiiis is 1, and whose eireumferenoe ia
denoted by C.
To compute C, i. e., 27t, and henee find the miniorieal
value of 7t approximatelj-.
Computation. 1. Inscribe a regular hexagon in the
given circle, and denote its side by &■
Then ySc = l. Art, 459.
.', perimeter of the inscribed hexagon = 6.
2. Inscribe a regular polygon of double the number
(12) sides.
Then, by the second formula of Art. 4GS,
Si2=V2- 1/4^^=0.51763809 +.
Denoting the perimeter of a regular inscribed polygon
of n sides by P„,
Pi2 = 12(0.5176380D+) = G.211t!5r08.
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MAXIMA AND MTKIMA
S. In the second formula of Art. 468, let h = 12.
.-. 834=^2—1/4— (0.51763809+)- = 0.26108238+, etc.
Hence Pw = 6.26525772.
Computing 8ig, P48, etc., in like manner, the following
results are obtained;
5.=l/2-
-Vi — i
= .51763P09.
= .2610523S.
= .130S0fi2^.
= .0Gr>43817.
= .03272346.
= .01030228.
= .00S18iaj.
■- Fit
: P2
•■. F,
.". Fa
.-. Pi
-■■ -Pt
■■• P-
= 6,21165708.
Su =1/2-
-i'4-(.5t76809)'
= 6.26525722.
^. =1/2-
-V4 — {.2til052;(S)'
= 6,27870041.
S„ =V'2-
-V4 — (.13080(i26)'
= 6.28206390.
S„.. = VT-
-V.i~-{.omi3sny
^ = 6.28290510.
-s,„=y'2-
-V4 — (.032733-46 J'
„ = 6. 28311544.
sn>=y'2^
-V 4 — 1.01636228)
6 = S.2S31604I,
Bj' continuing the computation it is found that the first
six decimal figures in the value of the perimeter of the in-
scribed polygon remain unchanged.
.-. C, or 271 = 6,283169 approximately.
.". 7t=3.14159 approximately,
MAXIMA AND MINIMA
470. Def, a maximum (see Art, 268) is the great-
est of a group of magnitudes, all of which satisfy certain
given conditions,
ThuB, the diameter is the maximum oliord wLicli imn be drawn in
471. Dep. a minimum 13 the smallest of a group of
magnitudes, all of which satisfy certain given conditions.
Thus, of all lines which can he dnira from a given point to ft given
Hue the perpendicular is the miniinum.
Certain masiiua and minima have already been studied, and we
now proceed to investigate more particularly those relating to regular
poljgons and tiie circle.
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Hi BOOK V. IT AXE GEOMETKY
Proposition XXI. Theorem
472. Of all trinngles which have two sides equal, tJutt
■i'liK/lr ill which these sides include a right angle is the
Given the A ABC and A'B'C in wliicli AB^A'B', OA
= C'A', and /.A' is a rt. Z.
To prove A A'B'C > A ABG.
Proof. In the A ACB, draw CD 1 AB.
Then CD < CA. (Why?)
.-. CD < CA'. Ax. 8.
But the &. ABC and A'B'C have equal bases. Hyp,
.■. these ^ are to each other as their altitudes, CD and
CA'. Art. 391.
.-. A A'B'C > A ABC.
(for C'J' > rii).
Q. E. B.
473. Iso perimetric figures are figures having equal
perimeters.
El. 1. Find the
iven parallel lines.
1 between two
Ex. 2. What is the largest stick that can be placed on a reetangu-
lar table 12 I 5 ft,, and not have an end projeuting over a side of
the table f
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MAXIMA. AND MIKIMA 287
Proposition XXII. Theorem
474. Of. all isopfrimetric Mangles which have the same
ase, the isoscelea triangle is the viaximitm.
I
Given the & ABO and AUT) having the same base AB
and eqnal perimeters, AC=CB, and AD and DB ub-
equal.
To prove A ACB > A ADB.
Proof. Produce AC to S, making CF=AG.
Draw FB. and from Z) as a center, with a radius equal
to T>B, describe an are catting FB produced in G.
DrawDG and AG.
Draw CS and DK 1 AB; also Gi and DP 1 FG.
Then Z A£f is a ri?bt Z , Art, 2G1.
( for it maij be inscribed in a semioirHe whose center is C and
s ACF).
Also ADG is not a straight line,
(/or, if it were, the A DAB and DBA WOiM be complements of the = A
DGB and DBG, respectwely, and hence icoiikl be equal, aitd .: A DAB
would be isosceles, which is contrary to the hypothesis).
:. .1F=AC+ CB=AD-\- DB = AD + DQ. Cocstr. Hyp.
But AD+ J)G > AG (Art. 02). :. AF > AG. Ax, 8.
.-. BF > BG. Art, 111.
.-. i r,F > h BG, or Cfl > KD. Ax. 10.
.-. A ACB > A ADB. .\yt.:m.
(J. E, O.
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288 book v. plane gkomktiiy
Proposition XSIII. Tueoeem
475. Of hopfirimetrin pohjgonx haviny the same number
of Hides, Ihf maximum is eqniUittral.
Given ABODE the maximum of all polygons having a
given perimeter, and a given number oE sides.
To prove -4 BGDB equilateral.
Proof. If AfiCDE is not equilateral, at least two of its
sides, as AB and BO, must be uneqnal,
If this is possible, on the diagonal -iOas a base, con-
struct a triangle having the same perimeter as ABC, and
having side AB'^B'C.
Then A AB'C > A ABC. An. 474.
To each of these uneqnals add the polygon ACDE.
:. AB'CDE > ABODE. As. a.
But this is contrary to the hypothesis that ABODE is
the maximum of the class of polygons considered.
Hence AB = BO, and ABODE is equilateral.
^___ "!- E. ».
Ei. Of all circles which are described on a given liue aa chord,
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maxima and minima 289
Proposition SSIV. Theorem
4'76. Of all polygons hmnng all sides given hut one, ihe
■maximum can he inscribed in a semicircle having the unde-
termined side as a diameter.
Given the polygon ABCDEF, (he maxitnnra of all poly-
gons having the sides BC, CD, BE, EF, FA, in common,
and J.B undetermined.
To prove that AB is the diameter of a semieircle in
■which ABCBEF Qs.x\ be inBcribsci.
Proof. Bravv lines from any vertex, E, to A and B.
The A BEA must be the maximum of ail A having the
sides BE and EA,
(for, if it is nnt, hj incrmsitig or decreasing the angle BEA, thf A KEA
can be chaitgeA till it is a maximum, ihe rest of figure, BCDE a'nl
EFA, laeaniime remmniug vnckanged; thus the area of the poly •
gon ABCDEF iBould 6e increased, wkich is contrary to Ihe
b'jpotliesis that ABCDEF is a raaxiiHum) .
.'. Z BEA is a right Z . Art 47^.
.". E is on the aemicircumference of whiuh AB is the
In like manner the other vertices, C, D and F, must lie
on the semicircumference which has AB for a diameter.
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BOOK V, TLASE GKOMETHY
Proposition XXV. Thbokkh
477. Of all polygons formed ;
that which can he inscribed in a ri,
iih the same given sides,
"If in the maxivium.
Given ABCDEa. polygon wliich cau be iuacribeii iu a O,
and A'B'CD'E' a polygon wliich has the same sides as
ABCBE, but which cannot be inscribed iu a O.
To prove ABGBE > A'B'G'iyB'.
Proof. From any vertex, A, of ABODE draw the
diameter AK, and join K to the adjacent vertices C and D,
Upon CD' constrnet the triangle C'E'D' equal to A CDK,
and draw A'ff .
Then area ABGK > area A'B'G'K'.
Also area AEDK > area A'E'D'K'.
Adding, ABGKVE > A'B'G'E'D'E'.
But A CED=A G'K'D'.
Subtracting, ABODE > A'B'CD'E'.
Art. ^76.
(Why t)
(Why?)
(Why T)
Q. E
478. Note. It might happen that one nf the parts of the soeond
figure formed by the diameter A'E', aa A'B'C'E', eould be ioEcribed ia
a Bemicircle, and .". = ABCK. How, then, mowid the above proof be
modified t
479. Cor. Of all isoperimftric polygons of a given
niimier of sides the maximum polygon is regular.
For it is equilateral (Art. 475), and can be inscribed in
a sircle (Art. 477), and is, therefore, rfifiiilar (Art, 417).
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MAXIMA AND MINIMA lifl
Proposition XXVI. Theorem
480. Of two isoperimetric regular pnhjrions, that ivhich
has the greater number of sides has (he greaUr area.
Given Jta regular polygon of any number of sides, as
three, and K' a regular polj'gon of one more, or four sides,
and let K and K' have equal perimeters.
To prove K' > K.
Proof, Prom any vertex, C, of K, ili-fiw a line CD to any
point D of the side AB, which meets one of the sides of
ZC.
Construct the A DCF, having CF^DA, and DF^CA.
:. A DCF= A CDA. Art. I01.
Adding A CBD, DFCB o K. A^. 2.
Henee the polygon DFCB has the same perimeter as K' ,
and the same area as K.
But DFCB is an ii-regukr, while K' is a regular poly-
gon of four sides.
.-. K' > DFCB. Art, 470.
.-. if > A'. Ax. 8.
In like manner it may be shown that a reguhir polygon
of one more, or five sides, >K, and so on,
Q. E. B.
481. Cor. Of isoperiineiric plane figures, (he circle is
tile maximum. That is, the area of a circle is greater than
the area of any polygon with equal peiimeter.
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292 BOOK V. FLAME GEOJIETEY
Proposition XXVII. Theorem
482. Of two equivalent regular polygons that wMch h
the less number of sides has the greater perimeter.
e'
, — p —
H
Given ^and E' two regular polygons having the same
area, and having their perimeters denoted by P and P', but
£7 having the less number of sides.
To prove I" > P.
Proof. Let ^ be a regular polygon having the same
number of sides as K', and the same perimeter as K.
Then K > S. Art. 480.
.-. JU > H. Ax. 8.
.-. P' > P. Art. 3P9.
El. 1. Find the area of a trij
and 12 in., and the im^ludifd Siiig
triangle having two aides of 6 an
Ex. 2. How long is the feni
g!e in which two of the sides are 6
is 90". Find the area of another
12 in,, and the inuluded angle 60".
about a garden 60x40 ft.? How
many square feet in the area of the garden t Find also the length of
fence and area of a garden 50 ft. square,
Ex. S. Find the area of aa equilateral triangle, a square, a heia=
gon, and a circle, in each of which the perimeter is 1 ft.
Ex. 4. Find the perimeter ot an eqnilateral triangle, a square, and
a circle, in each of which the area is 24 sq, in.
Sx. 6. What principle of ma^sima and minima is illustrated in
eaob ot the four preceding Exs.?
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483. Symmetry of polygons. Many of tlie propertiea
of regular figures can be obtained in a aimple and expedi-
tious way by the use of the ideas of symmetry.
484. An axis of symmetry is a line snch that, if part of
a figure be folded over upon it as an axis, the part folded
over will coincide with the remaining part of the figure.
EXERCISES. CROUP 46
Ex. 1. How many axes of symmetry liaa an
Ei, 2. How many lias a square? a rcgulai'
pentagon !
Ex. 3. How many lias a regular liesagoii ? a
rtgutar heptagon f
Ex. 4. How many haa a regular uctagon ! a
regular polygon of n sides T
Ex. 5. How many has a circle f
485. A center of symmetry for a polygon is a point
such tliiit jiiiv line di'awn throiigli the point and terminated
by the perimeter is bisected by the point.
squa
EXERCISES. CROUP 4
equilateral trlanglH a center
r pentagon a centei'
;.f syn
letry f Han a
iBtry ? Haa a
enter of sym-
Ex. 2. Has a n
regular hexagon ?
Ex. 3, 111 general, which regular polygons have a c
metry, and which do not ?
Ex. 4. Has a circle a center of symmetry t
Ex, 6. W'liich is the moat eymiuetiical plans figure atudied thus far f
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EOOK V. TLANE GIlOMliTiiY
SYMMETKY WITH RESPECT TO A LIME OK AXIS
486. Two points symmetrical -with respect to a line or
axis are poiuts such tliat the straight Hue joining ^ijem ig
bisected by the given liue at right angles.
Thus, if PP' is bisected by AB, and PP' is
X AB, the poiuts Pand P' are symmetrical
with respect to the axis AB.
487. A figure symmetrical with lespect
to an axis is a figure such that each point
in the one part of the figure has a point in
the other part symmetrical to the given
point, with respect to an axis.
488. Two figures symmetrical with respect -^f^^^-^-^^c
to an axis are two figures such that each point
in the one figure has a point in the other
figure symmetrical to the given point, with
respect to an axis.
li'
SYMMETRY WITH EESPECT TO A POITJT OR CEKTER
489. Two points symmetrical with respect to f^
a point or ceater are points such that the /
straight line joining them is bisected by the /
point or center.
Thus, if PP is bisected by the point (7, C is /
a center of synunetry, with respect to P and P. -y
490. A figure symmetrical with respect
to a point or center is a figure such that
each point in the figure has another point
in the figure symmetrical to the given
point with respect to the center.
49 1 . Two figures symmetrical with respect
to a center are figures such that each point in
one figure has a point in the other figure
symmetrical to it with respect to the center.
-^'^i?^^
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Proposition XXVIII. Theorem
492. If a figure is symmetrical with respect to tivo axes
tcMch are perpendicular to each other, it is symmetrical with
respect to their point of intersectimi as a center.
Given the figure ABO . . .H symmetrical -with respeot to
the two axes XX' and YY'; and XX' 1. YY' and intersect-
ing it at 0.
To prove ABC ... if symmetrical with respect to 0.
Proof. Take any point P in the perimeter of the
figure, and determine the points F' and P", symmetrical with
respect to P, by drawing FKP' L XX', and PLP" ± YY' .
Draw KL, P'O. OP".
Thon Pr [1 i'i", and PP" || XX'.
But PK^KP.
Also" Pfi:=and!i LO.
:. i'P = and || LO.
:. KLOP is a Cn , andA'i = and || PO.
In like manner it may be shown that KL = B.ih
.: PO=nnd II OP". Ax.
Art.
121.
Art
48fi.
Art.
157.
A
X. 1.
Art
160.
1 01
Art
122.
Hence any straight line drawn throngh 0, and terminated
by the perimeter, is bisected at 0,
. is a center of symmetry for the given figure. Ar!. 4Eio,
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I'LANE GEOMKTili"
EXERCISES. CRO
Ex. 1. The diagonals uf a regular pentagon
Ex. 2. 1( ABODE is a regular penta'
.rmnjrit..
■lUK a
Ex.3. Ill the same figure Bt' = irC.
Ex. 4. Also KCDE is a parallblograiii
El. 5. Also JC=JB + t!A-.
Ex. 6. Tlie diagonals of a regular p(
Ex. 7. The apotliei
lugle equals one-half
of a
Ex. 8. Tho altitude of au equilateral ti
equals one and a half times the radius of t
eumserjbed circle.
Ex. 10. The side of an equilateral triangle equals Ry"i, tl
oribed circle being denoted by It.
El. 11. Theapothom
hiilf the side of a r^uiar
, regular inscribed hexagon
rihed triangle { = -x]/'^)-
Ex. 12. The side of a reg
side of the regular inscribed triaogle.
Ex. 13, Find the side of an insc
the side of a circumscribed square.
1 of a
terms of E.
13 of the ins
rihed square in tarn
I inscribed square, i
ribed and circumscribed squares
one of these is double the other.
a of a regular inscribed triangle ia '-'-j — '
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ESEKCISES. THEOREMS Ivi
Ex. 17. Shovf that the area of a regular inscribed hexagon is
•^ — . What, then, ia the ratio of the area of a regular inaoribed tri-
angle to that of a regular iusi^ribed hexagon 1
Ex. 18. The area of a regular inscribed hexagon is three-fourtha
the area of the regular circumscribed hexagon.
Ex. 19. The area of a regular inscribed hexagon ia a mean pro-
portional between the areas of a regular inseribed and a regular cir-
cumsoribed triangle.
LSuo. On a figure similar to that of Prop. IV, p. 2(i6. let OP inter-
seet AB in K, and compare the A OKA, GAP, and OPA' .'[
"Ex. 20. The area of a regular inseribed polygon of 2n sides is a
mean proportional between the areas of regular inscribed and ciruum-
seribed polygons of n sides.
Ex. 21. The area of a regular inscribed oetagoa equals the area of
the rentaugle whose base and altitude are the sides of the oircum
scribed and inscribed squares respectively.
Ex. 23. An angle of a regular polygon is the supplement of the
angle at the center.
Ex. 24. Diagonals drawn froni a vertex of a regular polygon of n
sides divide the angle at that vertex into ?i^2 equal parts.
Ex. 25. The diagonals formed by joining the alternate vertices of
a regular hexagon form another regular hexagon. Find also the ratio
of the areas of the two heiagons.
Ex, 26. If squares be erected on the sides of a regular hessgon,
the lines joining their exterior vertices form a regular dodecagon.
Find also the area of this dodecagon in terms of 6, asideof the hexagon,
Ex. 27. The square of a side of an inscribed e^juilateral triangle
equals the square of a side of an inscribed square added to the square
of a side of an inscribed regular hexagon.
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NR CiEOMETliV
Ex. SO, Given AaBhC, AcB, BdC, somi-
eirnles; prove that the sum of the two cres-
cents AeBa and BdCb equals the ateaoE the
right triangle ABC,
Ex. 31. An equiangular poly)?oii in-
scribed ia a oircla is regular if the number o£ it
[SUQ. In the figure to Prop, III, p. 264, t
:. are AE^&sc BO, .". aide Jif^side BC, ete,]
sides be odd.
^ JEUC = ai-e EDOB
EXERCISES. GROUP M
JlASl.MA A^^D MINIMA
Ex. 2. Of equivalent piirallelogra
Ex, 3, Of i 30 peri metric ri
Ex. 4, Divide a giveu lir
■s whicli 13 t
Ex, 5. Find a point in the hypotenuse oi
the 8um of the squares of the perpendieular
the legs shall he a minimum.
Ek. 6. How shall a mile of wire fence
t triangle such thnt
3 from the point to
Ex. 7- Find the area in acres included by a mile of wire (enae if
it be stretched as a square, a regular hexagon, and a circle respectively.
Ex, 8, Of all triangles with the same base and equal altitudes, the
isosoelcs triangle has the least perimeter,
Ex. 9, Of all polygons of a given number ot aidta inscribed in a
given circle, the maximum is regular.
[SuLi. Prove the maximum polygon (1) equilateral, (3) equi-
angulaj,]
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EXEIJCTSES. SVMMETKY 299
E3. 10. Find tlie masimiuii leetatigle inscribea in a circle.
Ex. 11. Find the maximum rectaugle that eaii bo inscribed iu a
[Sue. Inscribe a square in tbe circle.]
Ex. 12. Of trapsKoida inscribed in a semicircle (liaviug the diam-
ter us one base), find the maximum.
Ex. 13. Divide a given straight line into two pnrts such that the
am of tiie aquarea ol these parts sliall be a minimum.
EXERCISES. CROUP SO
SYMMETRY
Ex. 1. A rhombus has how many axes oi! symmetry? Has it a
center of symmetry ?
Ex. 2. "What asls o£ symmetry has a quadrilateral which has two
pairs of equal adjacent sides f Has such a figure a center of symmetry ?
A parallelogram is symmetrical with respect to Ihe point
.1 — of its diagonals.
Ex. 4. A segment of a circle is symmetrical with respect to wha
xis!
Ex. 5. Has a trapezium a center of symmetry ! An axis o
Ex. 6. How many axes of symmetry have two equal circles take
as one figure 1 Have they a center of symmetry I
Ex. 7. "What axis of symmetry have any t*o circles f
Ex. 8. How must two equilateral triangles be placed so as to hav
& center of symmetry f So as to have an axis of symmetry f
Ex. 9. It two polygons are symmetrical with reference lo a een
ter, any two homologous sides are equal and [>arallel and drawn i
opposite direetioua.
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lOK V. I'LANE GEOMETUi
EXERCISES. GROUP 81
VARIOUS THEOREMS
if a eirtle equals 'wice tlie
an inscribod equilateral triangle la to the
Ex. 3. Tho diagoaal joining any two opposite voL'litts of a regulaf
Ex. 4. Tha radius of an inaeribed regulsL' polygon is a mean pro-
portional between its apotbem and the radius of tlie uireumacribad
regular polygon of the same number of sides.
Ex. 5. If the sides o£ a regular hexagon be produced, their points
of intersection are the Tortices of another regular hexagon. Also Had
the ratio of the areas of the two hexagons.
Ex. 6. Of all lines drawn through a given point within an angle,
and terminated by the sides of the angle, the line which is bisected at
the given point cuts off the minimum area.
Ex. 7. Eaeh angle of a regular polygon of n + 2 sides contains
_ rjgiit angles.
Ex. 8. The diagonals from a vertex of a regular polygon of ii -[- 3
sides divide the angle at that vertex into « equal parts.
Ex. 9. The sum of the perpendiculars drawn to the sides of a
regular polygon of n sides from any point within tho polygon e(|uals n
times the apothem.
Ei. 10. An equiangular polygon circumscribed about a circle is
[SUQ. Draw radii from the points of contact, and lines from the
vertices of the polygon to the center.]
Ex. 11. An equilateral polygon circumscribed about a polygon is
regular if the number of its sides is odd.
[Sua. See Figure of Prop- III, p. 264. Prove Jr=iiy, Draw radii
and prove /.T= IQ, etc.]
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EXEECISES. MISCELLANEOUS THEOREMS 301
Ex. 12, How must two equal isosceles triangles be placed so aa
to have a center of symmetry f How masl they be placed so as to
have an axiS of symmotry !
If, in a regular inscribed polygon of n sides, •%, denotes a side and r™
denotes the apothem, show thjil.
Ex. 13. For the triangle, S3 = Rv''3, r,, = i5.
Es. 14. For the square, S, = Bv'2, i-. = iJtv'3.
Ex. 15. For the heiagoa, S6 = K, r6 = iBv'3-
Ei. 18. For the octagon, Ss~RV'2—i/2.
Ex. 19. For the dodecagon, Su =• EV'2—t/Z.
Ex.30. Prove that V = K' + Sio'.
Ex.21. Trove that Si,' = S„' + S|f,'.
Ex. 22. IE ADB, AaC, CbB are semicir-
cles and DC ± AB, prove that the area
bounded by the three semicircumferencea
equals the area deaoribed on I>C as a diameter. ■^ ij Js
Ex. 23. If p„ denotes the perimeter of an iaseribed polygon of n
sides and P„ the perimeter of a oitonmaoribed polygon of n sideB,
1p^ P^
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302 BOOK V. I'LAKE GEOJIETIii
EXERCISES. CnOUP S2
PROBLEMS
Circumapribe about a ftivon eire5e
Ex. 1, An equilateral triangle.
Ex. 2. A square.
Ex. 3. A regular p':^ntagon.
Ex. 4. A regular hesagou.
Ek. 6. A regular dpcagon.
Ex. 7. Constpuet a regular pentagram, or .i-poinled star,
Ex. 8. Conatruet a he.Yagram, or O-pointed star.
Ex. 9. Constniet an S-pointed star.
Ex. 10. Construct iin angle of 36".
Ex. 11. Construct angles of 18', 9", 72°.
Ex. 12. Construct angles o£ '2i'', 12°, G", 48°.
eireumferenpe into two parts which shall
Ex. 14. Construct a regular pentagon which shall have twice the
perimeter of a given regular pentagon.
Ex. 16. Conatruet a regular pentagon whose perimeter shall equal
the sum of the parlmeters oC two given regular pentagons.
Ex. 16. CocBtruot a regular pentagon whose area sliail be twice
the area of a given regular pentagon,
Ex. 17. Construct a regular pentagon whose area shall be equal
to the sum of the areas of two given regular pentagons.
Ex. 18. Constru
oamferenee of a giv
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EXERCISES. PROBLEMS 303
Ex. 19. Construct a circle whose area shall be three timeH the
area of a given circle.
Ex. 20. Construpt a ciroumterenee equivalent ti the =iim and
another equivalout to the diBereneo, of two gi\en i irtmnleiencoa
] five equal parts by
Ex.
21.
Construe
a circle e
equivalent to th
e difference, o£ tv
Ex.
22.
Construe
a circle wh
ofagi
e„
irole
Ex.
23.
Bis
ct t!i
B iiroa of a
ference.
Ex.
24.
Divide th
e area of a
drawin
geo
ncen
rice
rcnmferene
Onag
ven
line
cons
ruct
Ex
25.
Ar
gular pentagon.
Ex
26.
Ar
egula
hexagon.
Ex
27.
Ai
egula
dodeeagon
Ex
28.
Ac
rcle
qiiivaleiit t
Ex
29
Ins
criLe
a regular oo
Ex
30
lus
cribe
a circle in a
Ex
31
Ins
eribe
a square in
mieircle.
;iven square.
Ex. 32. In a given equilateral triangle inscribe throe equal circlsa
each of which touches the other two circles and a side of the triangle
Ex. 33. In a given circle inscribe three equal circles which shal
touoh eaeb other and the given ci renin ferenoe.
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NUMERICAL APPTJCATTOXf! OF PLANE
OEOMETEY
METHODS OF KUMEEICAL COMPUTATIONS
493. Cancellation. In numerical work in geometry, aa
elsewhere, the labor of computations may frefjuentlj- be
economized. Those methods of abbreviating work, which
are particularly applicable in the ordinary numerical appli-
cations of geometry, may be briefly indicated, as follows:
To simplify numerical work by caneeliation, group
together as a whole all the numerical processes of a given
proilem, and make all possible cancellations before proceed-
ing to a final nuvterical redtiction.
Ex. YinA the ratio of tlie area of a rectangle, whose base and alti-
tude are 42 and 24 inches, to tlie area of a trapezoid, whoae bases are
21 and 35 and altitude 12.
Bj Arts. 383, 394 J
area of i-eetanRle ^ >2X 24 _ ^ ^fk?* ^r, j^,.
area of trapezoid 6(214-35) ^X^B
494. Use of radicals and of 71. Where radicals enter in
the course of the solution of a numerical problem, it fre-
quently saves labor not to extract the root of the radical till
the final answer is to be obtained.
Ei. 1. Find the area of a circle oiroumacribed about a squara
whose side is 8.
The diagonal of the square must be 8i/2 (Art. 34C).
.-. the radius of O =4i/2.
.'. by Art. 449, area of G = x(4i/2)' = 32ir = 100.6. Area.
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NUMERICAL COMPUTATIONS 305
Similarly in the use of tt, it frequently saves labor not
to substitute its numerical value for n tilt late in the process
of solution.
Ei. 2. Fiud the radius of a circle whose acefi is equal to the suia
of the areas of two circles whose radii are 6 and 8 inches, respectively,
Deaote the radius of the required eirele by x.
Then, by Art, 440, ^ j' = 36 tt + C4 ,r.
.-. Tl'= 100 TT.
,-. j' = 100, and x= 10, Fadiu.^.
495. Use of x, j, etc., as symbols for unknown quan-
tities. In some cases a numerical computation is greatly
facilifated by the use of a specific sijwhol for an imknoirti
quantity.
Ex. In a triangle whose aldea are 12, IS, and 2.'), find the segmeots
of the side 25 made by the bisector of the angle opposite.
Denote the required segments of Bide ^5 by
Then 12 : 18 = j: : 2;') — j:(Art. 332)
.-. 18j- = 12(2.j-j.-) (Art. 302)
.■.:r = 10.i
And 25~a: = lf..)" '
r Sc(jmfiits .
496. Limitations of numerical computations. Owing to
the limitations of human eyesight and of the instruments
used in making measurements, no measurement can be
accurate beyond the fifth or sixth figure; and in ordinary
work, stteh as is done by a carpenter, measurempnts are
not accurate beyond the third figure. As all numerical
applications of geometry are based on practical measure-
ments, it is not necessary to carry arithmetical work beyond
the fifth or sixth significant digit.
Other methods of facilitating numerical computations,
as by the use of logarithms, are beyond the scope of this
book.
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6Ub PLANE GEOMEIKY
KUMEBICAL PKOPEKTIES OF LINES
EXERCISES. CflOUP S3
THK KlfiHT THIANGLU
Ek. 1. Find Uie hypotenuse of a riglit lria»i;lo wht.so leE3 ave 13
and :i5.
Eji. 2. The hypotenuse of a right triangle i? 29. and one leg ia
20. Find the other leg,
Ei. 3. If a window ia 15 ft. from the (■round and the foot of a
ladder is to be 8 feet from the house, how long a. ladder is necessary
to reaeh the window P
Ex. 4. Find the diagonals of a rectangle whose sides are .j and 13.
Ex. 6. Find the diaconnl of a square whose side is I ft, 6 in,
Ex. 7. The diagonals of a rhombus .ire '2i and 10. Find a side,
Ex. 8. One side of a rhombua is 17 and one diagonal is 30. Find
tiie other diagonal.
Ex. 9. In a circle whose radiua is 5, find the length of the longest
and shortest chords through a point at a distance 3 from the center.
Ex. 10, In a circle whose radius is 25 in,, find the distance from
the center to a chord 48 in. long.
i in. fio
Ex. 12. A ladder 40 ft. long reaches a window 20 ft. high on one
Bide of a street and, if turned on its foot, reaches a window 30 It, high
on the other side. How wide ia tlie street J
Ex. 13. If one leg of a right triangle ia 10 aud the hypotenuse is
twit;e the other leg, hnd the hypotenuse.
Ex. 14. Find the altiludo of an equiiaterai triangle whose side isG,
Ex. 15, Find the aide of an equilateraltriaugle whose altitude is 3,
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NUMESICAL EXERCISER. LINES .307
Ex. 16. Find the side of & square whose diagonal ia 15.
Es. 17. One leg of a right triangle is 3, and the sum of tha
hypotenuBs and the other leg is 9. Find the sides.
Ex, 18. A tree SO ft. high ia broken off 40 ft. from the ground.
How tar from the foot o£ the tree will the top strilce V
Ex. 19, The radii of two oirolea are I and 6 in., and their oenters
are 13 in. apart. Find the length of the common esternal tangent.
Ex. 20. TiiG Bides of a triangle are 10, 11, 12. Find the length of
the projection of the side whose length is 10, on tlie side 12.
EXERCISES. CROUP S4
TRIANOJ-ES IN GENERAL
Ex, 1. The sides of a triangle are 13, 18 and 20. Find the aeg-
fflents of the side 20, made by the bisBOtor of the angle opposite.
Es. 2. In the same triangle, find the segments of the Bide 20, made
by the bisector of the exterior angle opposite.
Ex. S. If the legs of a right triangle are C and 8, find the hypote-
nuse, the altitude on the hypotenuse, and the projections of the legs
on the hypotenuse.
Ex. 4. Is a triangle acute, obtuse, or right, i£ the three sides are
5. 12, U; 5, 11, 12; 5, 12, 13; 4, 5, 6?
i are G, T and 8, compute the length
Et. 6. Also the length of the median on the same side.
Ex. 7. Also thu length of tlie bisector oE the angle opposite the
iiJe B.
Ex 8. It two Bides and a diagonal of a paiulloiotsram ftre 8, 13
^ia 10, find the other diagonal.
[SuG, Uao Art. 352.]
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PLAKE GEOMETRY
El. 11, The hypoteniisB of a right triangle 'in 10, and the altitude
on the hjpotenuaa is 4. Find tho aegments of the hypotenuse and tlie
legs.
El. 12. Find the three medians, the fliree bisectora, and the tLreu
altitudes ol a triangle whose sides are 13, 14, 15,
EXERCISES. CROUP SS
CIRCTTMFRRENCF.S AM> ARCS
T. - 55
Lamg fl-^.-yt
Ex. i. Find thn ciroumEerence ot a eirclf whose radius in I ft. 9 in.
Ex 2. Find the radiuB of a circle whose Rireumferenee is 121 ft.
Ex. 3. A biojele wheel 28 in. in diameter makes, in an afternoon,
3,000 revolutions. How far does the bicycle travel F
Ex. 4. What is the diameter of a wheel whkh makes 1,400 revolu-
tions in going B,800 yds. ?
Ez. 5. If the diameter of a circle is 20, find the length of an arc of
60°; also of 83°.
Ex. 6. If the length of an arc is 14 and the radiua is G, End the
namber of degrees in the arc.
Ex. 7. If the arc of a quadraut m 1 ft. in length, find the diameter.
Bk. 8, Two eonoentrie circumferences are 88 and 132 in. in length,
respectively. Find the width of the Bircular ring between them.
Ex. 9. If the year be taken aa 365i da., and the earth's orbit a
circle whose radiua is 93,250,000 miles, find the velocity of the earth
in its orbit per second.
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NUMERICAL EXERCISES. LINES 309
Find the r&dius and circumference of a circle circumscribed about
Ex, 10. A square whose aide is 5.
Ex. 11. An equilateral triangle nhose side is 4.
Ex. 12. A rectangle whose sides are 12 and 5,
Ex. 14. Find the diameter of a cirt
;le eircH
angle whose sides aro 7, 15, and 20,
Ex. 15. Find the radius of a oirclf
i whose
the perimeter ol a square whose diagoni
il is lU.
EXERCISES. CROUP liS
CBOKDS, TANGENTS, AND SECANTS
ES. 1. Two intersecting chords of a cirule are 11 and 14 in., and
the segments of the first chord are 8 and 3 in. Find the segments of
the second chord.
[St;G. Denote the required segments bf :c and 14 — e,]
Ex. 2. In a circle whose radius is 12 in., a chord 16 In. long is
passed through a point 9 in, from the center. Find the segtnents of
the chord.
Ex. S. Two secants drawn from a point to a circle are 24 and 37
in. long. If the external segment of the first is G in., find the external
segment of the second.
Ex. 4. From a given point a seeant whoso external and internal
segments are 9 and 16 is drawn to a circle. Find the length of the
tangent drawn from the same point to the circle.
Ex. 5. Frenj a given point a tangent 24 in. long is drawn to a
circle whose radius ia 18 in. Find the distance of the point from the
center.
Ex. 6. I( a diameter GO in. long is divided into 5 equal parts by
«l»ords perpendicular to it, find the lungth of tiu) chowlB,
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PLANE GEOMETKY
Ex. 8. If the earlh is
will tbe light of a HgLlhou
EXERCISES. CnOtIP Bt
LINES IN' SLMILAli FICURiiS
Ex. 1. K the Hides of a triiwgle are fi, 7 and S, and the shortest
Bide of a similar triangle is 18, find the other aidea of tbe secoLd
Ex. 2. If a post 5 ft. high easts a shadow :! ft. lotij;, find the
height of a ateeple wkiph easts a shadow 90 I't. long.
Ex. 3. In a triangle whose base is 14 and aliitude 1!!, a line is
drawD parallel to the base and at a distauee 3 froni the base. Find
tbe length of tbe line thus drawn.
Ex. 4. The upper and lower bases of a trapezoid are 13 and %
and tbe altitude ia H. If the legs are produced till they meet, find
the altitnde of eaph of tbe two trianglts thus formed.
El. 5. If the upper .-ind lower bases of a trapezoid are i, and li,
and the altitude ia h, find the altitude of each of tbe triangles formed
by producing the legs.
Ex. 7. If the perimeter of a legular poljgon is three times the
the ratio of their apoihema t
Es. 8. It the eireumterenees of two eirelea are 600 and 400 ft.,
wbat is the ratio of their diameters 1
Ex, 9. In the preceding example, if a ohord of the first circle ia
30, what is the length of a chord in the second circle, subtending
the same number of degrees of arc 1
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NUMERICAL EXEECISES, AREAS 6
COMPUTATIOW OF AREAS
EXERCISES. CROUP iiS
AREAS OF TRIANGLES
i of a triangular field whose b
El. 2. Find the area oC a triangle wlioae aldea are 10, 17, and
El. 3. Find the area in aerts ot a liuld whose sides are (iO,
ud no chains.
d of a triaag-ulur tield t^xch i>S «li
clialcs.
Find the area of
Ex. 5. Ad isosaelea triangle whose base \3 16, and eai^li of wh
legs la 34.
Ex. 6. Aa equilateral triangle whoEe altitude is 8.
Ei. 7. A right triangle in which the segments of tha hjpotec
made by tlie altitude upon it are 12 and 3; also, in one in which
Ex. 8. An isosceles right triangle whose hj-potennse is 12.
Ex. 9. A right triangle in which fhe hypotenuse is 41 and one leg
is 9.
Ex. 10. Find in two ways ibe area of a triangle whose sides are
6, 5, 5.
Ex. II. A side of a given equilateral triangle is 4 ft. longer than
the altitude. Find the area of the triangle.
Ex. 12. The area of an isosceles triangle is 144 and a leg is 24.
Find the base.
Ex. 13. The area of an equilateral triangle is 4t/3. Find a side.
Ex. 14. The area ot a triani;ie is 1125, and if.h: f = 2 : ;i : 4. Find
Ex. 15. The area o£ a triangle is G sq. in., and two ot its aides are
il aud 5 in. Find the remaining eida.
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i'LANE GEOMETHY
EXERCISES. GROUP SB
AREAS OP OTHER RECTILINEAIE t
■ir.URES
PimJ tlie area of
Ex. 1. A par«.lleloi;ram whosa Ijaso is 2i ft. fi '
tuaei8l2ft. Oiu.
ill. and w
Ex. 2. A trapt^zoia whose uasvs are 12 ami 20
tude is n ft.
in. and w
Ex. 3. A vhombus whoae diagonals are 9 ft. and 2 yds.
Ex. 4. A quaJrilateral in wliieh the sides All, BC, CD, DA a
13, 14, 15 and the diagoual AC is 17.
Ei. 5. A quadrilateral iu which the sides ai-o £7, 36, 30, 2
the angle included betweeu the first two siciea is a right angle.
Eit. 6. A square whose diagonal is 12 in.
Ex. 7. Piud the number of boards, ea<;li 4 yds. long and
wide, which are neeessarj to cover a floor 48 X 24 ft.
Ex. 10. A rectangular garden contains 4,524 aq. yds. and 13 20
yds. longer than wide. Find its dimensions.
Ex. 1 1. Each Bide of a rhombus is 24 ft. and each of the larger
angles 13 double a smaller one. Find the area.
Ex. 12. Find the area of a rhombus one of whose sides is 17, and
one oC whosB diagonalB Is 3U.
Ex. 13, The area of a trapeaoid is 4 acres, one base Is J2U yds.,
and the altitude is 100 yds. Find the other base.
Ex. 14. The bases of an iaoseeles trapei^oid are 20 and 30 and the
legs an IT. Find the area.
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NUMERICAL EXEKCISES. AREAS 313
Ex. 15. The base of a triangle !a 20 and tbe altitude 13. Find
the length of a line parallel to the base ivhieh cute off a trapezoid
vi\ioRe area is 80 sq. £t,
[Suo, Denote the altitude ot trapezoid by 18— a; and Gad its
•■ppBr base by similar triangles.]
Ex, 16. The perimeter of a polygon, circumscribed about a circle
trhose radius is 20, is 340, Find the area of the polygon.
Ex. 17. The area of a rectangle is 144 and the base is three times
the altitude. Find the dimensions.
Ex. 18- Find theareaof aregularhesagon oneof whose sidesialO.
Ex. 19. Find the area of a regular deeagou inscribed in a eircie
whose radius is '20.
Ex. 20. f iud a aide of a. regular hesagoii whose area is 200 sq, in.
EXERCISES. CROU? 60
AREAS OF CIRCULAR FIGURES
Ex. 1. Find the area in aeres of a circle whose radius is 100 yds,
Ex. 2. Find the radius in inches o! a circle whose area is 1 sq. yd.
Ex. 3. Find the area ot a circle whose circumference is j).
Ex. 5. Find the radius of a circle whose area equals
^e areas of three circles whose radii are 20, 2H, 10.
Ex. 6. In a eircie of radius 50 find tbe area of a sector
Ex. 8, IQ a circle whose radius is 7, the area of a sector is 43 sq. ft.
Find the number of degrees in its angle.
Kx. 9. Ill a circle whoso radius is 10, fiud the sum ot the segmeiitB
formed by au inseribed square.
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PLANE GEOMETHV
Ex. li. The same pond is surrounded by a driveway 30 ft. wide.
Find the area of the driveway.
Ex. 12. Two tangents to a nirele, whose radius is 15, include an
angle of 60°. Find the waa Included between the tangents and the
radii to the points of contact.
Ex. 13. Find the length of the tether by which a cow must be
tied, in order to graze over eiiactly one acre.
Ex, 14, Three equal eii'cles touuh each other externally. Show
that the area included between them iaJtHyS — ^)-
Kx, 15, If the area inclnded between three equal circles which
touch each other externally is a square toot, find tlie radius of each
EXERCISES. CROUP ei
AREAS OF SIMll.AK FIGURES
Ex. 1. The homoloKons sides of two similar triangles are 3 and 5.
Find the ratio of their areas.
Ex, 2. The liomologoua sides of two similar polygons are 4 and 7,
and the area of the first polygon is 112, Find tlie area of the second
polygon,
Ex.3. The radius of a circle is e. Find rlie radins ot a circle hav-
ing three times the area of the given circle.
Ex, 4. The areas of two circles are as IC to p, and the radius of the
first is 8. Find the radiua of the second.
Ex. 6. The sides of a triangle are 5, G, 7. Find the aides of a
similar triangle containing 9 times the area of the given triangle.
Ex. 6. If, ia finding the area of a circle, a student uaeB 1> = 50 as
K = 50, how will the area as computed differ from the correct area 1
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MISCELLANEOUS NUMERICAL EXERCISES
Et. 7. In a triangte whose base \a 24 in. and altitude is 18 ii
ftltituiie is bisected by a line purallel to the base. Find tbe a
the triangle cut off.
Ex. 9, In a circle wboBe diameter is 30 in,, what are the diameters
of ooncentric eireumferences which divide the area into three equiva-
lent parts T
Ex. 10. If a circle be eoostrueted on the ridius o! a given circle,
Hud segments, one in each eirGle, be formed by a line drawn from the
point ol contact, find the ratio of the ,
EXERCISES. CROUP 63
OENERAL NUMERICAL EXERCISES EH PLANE OEOMETRY
iiid the base is IQ.
e equivalent to this triangle.
Ex. 4. The Rides of a triangle arc 7, 8 and S inches. Find the
aides of a triangle of four times the area. Also, of twice the area.
Ex. 5, The temple o! Herod is said to have aooommodated 210,000
people at one time. If each person required 27 X IS in., and one-third
the space iiiaide the temple be allowed for walls, sanctuaries, etc.,
what were the dimensions of the temple, if it was a square f
mgle are 12, 16 and 21., vrhat are the
/ the bisector of the angle opposite t
lateral in order
m aiif;le of (10°,
Find the diameter o£ a wheel which, in a mile, maires 480
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PLASE GEOMETKV
Find the radius of a circle equivalent to
Ex. 10. A square mlioae side is 10.
Ei. II, An eijiiilateral triangle whose siile Is 12.
Ex. 12. A trapezoid whose bases aro IG and 18 and altitude 9.
Ex. 13. A aemicirelo whoso raijins is 15,
'!■ of ft circle whose area shall be e<]uiTa-
I whose diameters are U4 and 17.
Ek. 15. A oirele, a square, and an equilateral triangle each, have
s perimeter of 12 yds. Find the area of eai^h figurii.
Ex. 16. In a circle whose area is 400, the area of a seetor is 125.
Find the angle of the sector.
Ex. 17. How many acres are included w
track, if the traek is in the shape of a rei
it IE wide 1
Ex. 19. Find the area of the circle circumscribed about a
angle whose sides are 40 and 9,
Ex. 20. One leg of a right triangle
teen the hypotenuse and the other leg
Ex. 21. Find the area of an isosceles
Ex. 22. In a triangle whose sides are 16, 18, 20, find the length of
the altitude, median, and bisector of the angle opposite the longest
Ex. 23. A line 16 inches long is divided internally in the ratio of
3:5; find the segments. Also find the segments when the line is
divided externally in the same ratio.
Ex. 24. A line 16 Inches long is divided in extreme and mea&
ratio. Find the segments.
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MISCELLANEOUS NOMEEICAI, EXERCISES 317
Ex, 25. If a line is divided in extreme and mean ratio and the
Bmaller segment ia 4, flnd tlio whole line.
Ex. 28. In a circle whose diameter is 20, a chord is passed throuKh
a point at a distance 6 from the center, perpendicular to the diameter
through that point. Find the length ot this chord, and ot the chords
drawn from its extremities to the ends of the diameter.
En. 29. Eiioh Itg of an isosceles trapezoid is Jll, and onH base
exceeds the other by 16. Find the altitude.
Ex, 30. It three arcs, each of 60° and having 10 for a radius, are
to the other two, find the area included by them.
Ex. 31. Find the area of a trap?
izoid whose legs i
ire 4 and 5, and
whose bases are 8 and 11.
Ex. 32. A square piece o£ land :
and a<-irpnlar pi,
ice each contain
1 acre, llow many more feet of fenc
does one require
. than the other ?
El. 33. IE the base of a triangle is doubled and the altitude re-
mains unchanged, how is the area affected ! If the altitude is doubled
and the base remains unchanged t It both the base and the altitude
are doubled f
ot
Ex. 37. Find the side of an equilateral triangle equivalent to a
oitcle whose diameter is 10.
Ex. 38. The area of a rhombus ia loti sq. in., and one side is 1 ft.
1 in. Find the diagnnuls.
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318 PLAN'E GEOMETRY
Ex. 39. Tlie EJiipB of atriansle are S, in, 12, Find the aicis of
the triangles iii.icle by llie bisector of the angle opposite Ihu side 12.
Ex. 40. In a circle of area 275 sq. ft., a re<^tangie of area 150 aq. ft.
is inaeribed. Sliow how to find the sides of the reetaogle.
EXERCI3E3. CROUP 63
EXERCISKS INVOLVING . THE METRIC SYSTEM
Ex. 1, Find the area of a triangle of which the baee is 16 dm. and
the altitude SO cui.
Ex. 2. Find the area of a triangle whofe sides are 6 m., TO dm.,
800 cm.
Ex. 3. Find the area in square meters of a circle whose radius is
14 dm,
Ex. 4. If the hypotenuse of a right triangle is 17 dm, and one log
iB 150 cm., find the other leg and the area.
Ex. 5. It the circumference of a circle \i 1 m,, find tlio area of
the circle iji square deeimeters.
Ex. 6. Find the area in heetares, and also in acres, of a circle
whose radius is 100 m,
Ex. 7, If the diagonal of a rectangle is 35 dm, and one side is 800
mm., find the area in square meters, and also in square inches.
Ex, 8, Find the area of a trapezoid wiiose bases are 600 em. and
2 m., and whose altitude is SO dm.
Ex, 10. In a given circle two chorda, whose lengths are 15 dm.
and 13 dm,, intersect. 3f the segments of the first ehord are 12 dm.
and 3 dm., find the segments of the second chord,
Ex. 11. Find in det imetera the radius of a circie equivalent to a
square wbosa side is 1 ft. 6 in.
Ex. 12. Find in feet the diameter of a wheel which, in going 10
kilometers, makes 5,000 revolutions.
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SOLID GEOMETRY
Book VI
LINES, PLANES AND ANGLES IN SPACE
DEFimTIONS AND FIRST PRINCIPLES
497. Solid Geometry treats of the properties of space of
three dimensions.
Many of the properties of space of three dimensiona are determined
by use of the plane and of the pcopertiea of plaue figures already
obtained in Plane Geometry.
498. A plane is a surface such that, if any two points
in it be joined by a straight line, the line lies wholly in the
surface.
499. A plane is determined by given points or lines, if
no other plane can pass through the given points or lines
"without coinciding with the given plane.
500. Fundamental property of a plane in apace. A
plane is determined by any three points not in a straight line.
For, if through a line con- ,c
necting two given poiuts, A
and B, a plane be passed, the
plane, if rotated, can pass
through a third given point,
C, in but one position. ■'*
The importance of the above principle is seen from the faet that it
reduces an unlimited surface to three points, thus making a vast
econoijij to the attention. It also enables us to eouuect different
planes, and treat of their properties aystematically.
(3iy)
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320 BOOK VI. SOLID GEOMETRY
501 . Other modes of determining a plane. A plane may
also be delerm'uied by any equivalent of three points not iu
a straight line, as by
a straight line and a point outside the, line; or hy
two intersecting straight lines; or by
two parallel straight lines.
It is often more eonvenient to use one o£ these latter raethods of
determining a plane than to reduea the data to tliree points md use
502. Representation of a plane in geometric figures. In
reasoning concevuiug the plane, it is often an advantage
to have the plane represented in all directions. Hence, in
drawing a geometric fignre, a plane is usually represented
to the eye by a small parallelogram.
This ia virtually a donble aae of two intersecting lines, or of two
parallel lines, to determine a plane (Art. 501].
503. Postulate of Solid Geometry. The principle of
Art. 499 may also be stated as a postulate, thus:
Through any three points not in a straight line ior their
eqnivaJeni) a plane may be passed.
504. The foot of a line is the point in which the line
intersects a given plane.
505. A straight line perpendicular to a plane is a line
perpendicular to every line in the plane drawn through its
foot.
A straight line perpendicular to a plane is sometimes called a normal
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LINES AND PLANES 321
506. A parallel straight line aud plane are a Hue aad
plane which cannot meet, however far they he produced.
507. Parallel planes are planes wliich caonot meet,
however far they he produced,
508. Properties of planes inferred immediately.
1. A atraighl line, not in u yircti plane, can in!erseci IJie
given plane in but one point.
For, if the line intersect the given plane in two or more
points, by definition of a plane, the line must lie in the
plane. Art. 498.
2. The intersection of two planes is a straight line.
For, if two points common to the two planes be joined
by a straight line, this line lies in each plane (Art. 498) ;
and no other point can be common to the two planes, for,
throagh a straight line and a point outside of it only one
plane can be passed. Art. soi.
Ex, 1, Give an esample o£ a plane Burfa<ie; of a eiu'vod siirfaee ;
of a Burfaee, part plane and part curved; of a surface composed of
different plane Burfaees.
Ex. 2, Four points, not all in the same plane determine how
many different places ? how many different straight lines ?
Ex. S. Three parallel straight lines, not in the same plans, deter-
mine how many diHerent planes ?
Ex. 4. Four parallel straight lines can determine how many difter-
Ex, 5. Two iiilerseetins straight lines an.! a point, not in thei!,'
plane, determim; how many difie-rent planes 1
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322 BOOK VI. tiorjD geometky
Propositio>- I. Theokem
509. If a siraiglht line is perpendicular io each of ttvs
other straight lines at their point of intersection, it is per-
pendicular to the plane of those lines.
Given AB J. lines BC and BD, and the plane UN pass-
ing through BC and BD.
To prove AB ± plane MN.
Proof. Through B draw BG, any other line in the plane
MX.
Draw any couveuieut line CD interseeting BC, BG and
BB in the points C, G and B, respectively.
Produce the line AB to F, making BF^^AB.
Connect the points C, G, D with A, and also with F.
Then, in the A ACD and FCD, CD=CD. Ident.
AC- CF, and AD^BF. Art. 112.
.-. AACB^AFCB. (Whyf)
.-. / ACB= I FCD. (Why!)
Then, in the A ACG and FCG, G6=CG, (WbyT)
AC^CF, and / ACG = IFCG. (Why!)
.-. A AGG^AFCG. (Whyt)
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LINES AND PLANES 323
.-. AO=GF. (Why?)
. B and 6 are each equidistant from the points A and F.
:. BO is X AF; that is, AB ± BG. Art. ii3.
.-. AB X plane jl/iV, Art. 505.
(forii is X any line, BG, in the plane MJS, through its fool).
q. E. D.
PROrOSITION IT. TlIEOItKM
510. AU tiie perpeitdieulars that can be tirawn to a
given line at a given point in the line lie in a plane perpen-
dicular to the line at tJte given point.
X
■<ri\.
Given the plane M]V and the line 5(7 both X line JBa,i
the point B.
To prove that BG lies in the plane MI^.
Proof. Pass a plane AF through the intersectinff lines
AB and BC. Art. 503.
This plane will intersect the plane MN in a straight
line BF. Ari. 508, 2.
Bat AB X plane MN (Hjp.) .". AB X BF. Aft. 505.
Also AB X BC. Hyp.
.-. in the plane AF, BC and BF X AB at B.
/. BG and BF coincide. Art. 71.
But BF is in the plane MN.
.: BC must be in the plane 31N,
{for B(J uMWiklcs wUh BF, ichidi. ties m the jilaiic MX).
e, E. B.
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824 BOOK VT. SOTJD GFXl^ilETliY
611. CoE. 1, At a given point Bin iliestm'ijhl line AB,
io construct a plane, perpendicular to the line AB. Pass a
plane A-F through AB in any convenient diveiiliou, and iq
the plane AF at the point B coiiBtruet BF ± AB (Art.
274). Pass another plane through AB, and in it construct
BP X AB. Through the lines
BF and BP pass the plane
MX (Art. 503). MX is the f
required plane (Art, 509).
512. Cor. 2. Through a
given external point, P, to pnsa
a plane perpendicular to a
given line, AB. Pass a plane through AB and P (Art.
503) , and in this plane draw PB ± AB (Art. 273) . Pass
another plane through A B, as AF, and in AF draw BF X
AB at B (Art. 274). Pass a plane through BP and BF
(Art. 503). This will he the plane required (Art. 509).
513, Cor. 3. Through a given point hut one plane can
be passed perpendicular to a given line.
Ex. 1. Five points, no tour of which are in the same plaoe, deter-
line how many different planes ! how many different straight lines 1
A atvaight line and two points, not
le, determine how macy difiecent plat
Ex. S. In tte figure
GDF are equal.
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LINES AND PLANES
Proposition III. Problem
514. At a given point m a plane, to erect a perpendicu-
lar to the plane.
1 ''• 1
sz^^^
\-^v i\
Given the point A in piano MN.
To construct a line pei-pendicular to MN at the point A.
Construction. Through the point A draw any line CD in
the plane MF.
Also through the point A pass the plane PQ X CD
(Art. 511), intersecting the plane J/jV iu the lineiSS. Art, 508, 2.
In the plane FQ draw AK 1 line ES at A. Art. 274.
Then AK is thi; ± requirud.
Proof. CD ± plane Py. Coiistr.
;. CD 1 ,4/t:. Art. 505.
Henw AK X CD.
But il/r X 7i\S\ Coustr.
.-. AK X phine MN'. Art. 5oy.
Q. E. F.
615. CoK. .(l( fl given point in a plane but one perpen-
dicular to the plane can he drawn. For, if two X could be
drawn at tlie given point, a plane conld be passed through
them intersecting the given plane. Then the two X would
be in the new plane and X to the same line (the line of
intersection of llie two planes, Art. 505); which is im-
possible (Art. 71).
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■rlh BOOK VI. fiOLID CEOJIETilY
Proposition IV. Problem
516. Fimn a given point iviiliout a jilatie, to draic a line
pcrpfndicifhii- to the plane.
Given the plane MX and the point A external to it.
To construct from A a line ± plane MX.
Construction. In the plane MX draw any convenient
iine BG. Pass a plane through BC and A (Art. 503), and
111 this plane draw AD ± BG. Art. 273.
In the plane MX draw LD X BC. Art. 274.
Pass a plane through AD and LD (Art. 503), and in
that plane draw AL A. LD. Art. 273,
Then AL is the ± required.
Proof. Take any point C in BG except X>, and draw LO
uud A G.
Then &. ADG, ADD and LDG are right A . Consti
.-. J6''=Zd^ + DG'. Art. 400.
.-. AC^ = ZL^ + LD' -f 'DC'. Art. 400, Ax. 8.
.'. AC^^AL^ -{■ LG'^. Art, 400, As. 8.
.-. ZAiCisaright /. . Art. 351,
But AL ± LD. Conatr.
.-. AL X MX. Art
517. Cor. Bitt one perpendicular can be dratcn from a
given external point ia a given plane,
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LINES AND PLANES 327
Proposition V. Theorem
518. I. Oblique lines drawn from a point !o a plane,
meeting the plane at equal distances from the foot of the
perpendicular, are equal;
II. Of ftco oblique lines drawn from a point to a plane,
but meeting the plane at unequal dlniances from the fool of
the perpendicular, the more remote is the greater.
Given AB ± plane MN, BD=BC, and BH > EC.
To prove AD^AC, and AM > AC.
Proof. I. In the right A ABD and ABO,
AB = AB, and BD = BC. (Why?)
.-. AABZ>=A^£0. (Why')
.-. AI> = AC. (Wby?)
II. On Offtake iSJP^Be, and draw AF.
Then AF^AC (lypart of theorrm jin^t prm'f,d) .
But AIT > AF. (Why?)
.-. AU > AC. As. 8.
Q. E. D.
519. Cob. 1. Conversely: Equal oblique lines drawn
from a point to a plane meet the plane at equal distances
from the foot of the perpendicular drutni from the same point
to the plane; and, of two unequal li}iex so drawn , the greater
line meets the plane at the greater distance from the foot of
the 'perpendicular.
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328
SOLID r.EOJlETl'.Y
520. r'dJi 2. The locus of a point in space cijuklhiant
from all llif points in the circumference of a circle is it
ulraight line passing through the center of the circle and
perpendicnhtr to its plane.
521. Cor. 3. The perpendicular is the shorlpst line Ihut
can be drawn from a gii-en point to a given plane.
522. ]>KF. The distance from a point tu a plane is the
perpendleiihir drawn from the point to the jilniit;.
Proposition VI. Theorem
523. If from the foot of a perpendicular to a plane a
line be drau-n at right angles to any li7te in the plane, the line
drawn from thtpoint of intersection so formed to any point in
the perpendicular, is perpendicular to the line of the plane.
Given A7i ± plane MX. and BF X CJ), nny Hue in 3LY.
To prove AF J_ CJJ.
Proof. On €1) take FP and FQ equal segments.
Draw A P, HP, AQ, BQ.
Then BP^BQ. Art. 112.
Hence AP^AQ. Art. ma.
.'. io the line AF, the point A is equidistant from P
and Q, and F is equidistant from P and Q.
.: AF ± €T). (Why!)
Q. E. ».
Ejc, In the above figure, it Ali^G, AF^S, and jy = lU, find QF,
DF and C^.
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LIKES AND I'LASES
Proposition VII. Theorem
524. Two straight lines perpetnUcitlar to ike same plane
>■€ parallel.
i
\
/
\
•f /
/ """"T I
Given the lines All ami CD ± plane ilW.
To prove AB 11 CD.
Proo(. Draw BI), anil Uu-ousli J), in the plane MS,
draw FR X BI>.
Draw Ai).
Then
;j/< J- FII. Con.lr
Ai> ± FM. Art. 023.
0/' i F7Z. Art. 505.
.-. BB, AD and (7/") rn-p all ± J'J/ at the poiut T>.
.". R/', A7' and Cf.) all lie in the same plane. Art. 5io.
.-. AB imd CD ave iu the same plan'^. (Why?)
But Ali and (JIf are ± B'li. Art. 50o.
.-. AB and 6'Z> are 1|, Art. vi\.
t). E. D.
525. Cor. 1. If one of tiro parallel liiie^ j.v pn-pendicn-
iar to a picnic, the other is perpendicular to the plane also.
For. if AB and CI) be ||, and AB ±
plane PQ, a line drawn from C L TQ 4 f
must be II AB. Art- ;i?4.
But CP iiuist coincide with tin? line V ^ ;)\
so drawn (Art. 47, 3); .'. CD 1 7'(^ ^ —''^^
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:^,:";o
SOLID GEOMKTP.V
526. Cor. 2. Ifin-o straight lines are each parallel to
a third straight Uhp, tJiey are parallel to each other. For,
if a plane be drawn 1 tu llii? 1;liird line, each of the two
other Sines must be 1. to it {Art. 525), and therefore be 1|
to each other (Art. 5-4).
y*iioi'OfiTio\ YIII, Theorem
527. If a sirairjhi line ex/erna! to a given plane is paral-
lel lo a line in the plane, then ihejirst line is parallel to the
given plane.
B Cn in the plane MN.
!■ .¥_V.
Given tlio straight linpyl/; II
To prove Ml \\ plai
Proof. Pass a plane thron^h the || lines AB and CD.
If AB meets MN it must meet it in the line CT).
But AB and CD cannot meet, for they are ||. Art, 120,
.". AB and MN cannot meet and are parallel. Art. 506,
Q. E. 9.
528. Cor. 1. // a straight line is parallel to a plane,
the intersection of the plane with any plane passing through
the given line is parallel to the given line.
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LINES AND PLANER
529. Cor. 2. Through a given line ^_ ,g
(CD) to pass a plane parallel to another q
given line (AB). q..^i^^,^. r
Through P, any point in CD, draw ^
QR |[ AB (Art. 279). Througli CDaui
QB pass a plane (Art. 503). Tliis will be tlie plane
reqaired (Art, 527).
Jt AB and CD are uot parallel, but one pluiie can be
drawn through CD\\AB.
Proposition IX. Theorem
530. Two planes perpendicular to the sanu
re parallel.
Given the planes MN and PQ X line A 11.
To prove MN \] PQ.
Proof. If MN and PQ a.ve not parallel, on being pro-
duced they will meet.
We shall then have two planes drawn from a point per-
pendicular to a given line, which is impossible. Art. 5i3.
.■, MN and PQ are parallel. Art. 507.
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f;oLin (";En:MET];v
Proposition X. Thk(jr!:m
631. // two parallel planes are cut by a tJih-d plane, ike
intersections are parallel lines.
Given M'N' and PQ two II planes iiitersecteil by the planuj
BS in tbe lines AB and CD.
To prove AB \\ CI).
Proof. AB and CT) lie in the same plane US.
Also AB and CI) panKOt meet; for if they did meet
the planes ilO' and FQ would meet, which is impossible.
Art- r>07.
.-. AB and ('/> are pm-illoi. Art. 4L
0- E. D.
532. Cor. 1. Parallel lines included between parallel
planes are equal. For, if AC and BD are two parallel lines,
a plane may be passed through them (Art. 503), intersect-
ing MN and BQ in the 11 lines AB and CD. Art. 531.
.". ABDC is a pariillelogram. Art. U7.
.-. .1(7= CD, Art. 155.
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LISES ASD PLANES 333
533. Cor. 2. Two parallel planes are everywhere equi-
distani .
For lines 1 to oae of them are il (Art. 524). Hence
the segments of these lines included between the || planes
are equal (Art. 532).
Proposition XI. Theorem
534. If two infersecting lines are each parallel to a given
plane, the plane of these lines is parallel to tJie given plane.
\ :
>
1^ \.
'\
,.■-'1
"■J \
Given the lines AB and Cl>. eatih \\ plane PQ, and inter-
secting in the point F: and lilN a plane through AB and
CD,
To prove MX \\ PQ.
Proof. From the point F draw FS X PQ.
Pass a plane through FC and Fff, intersecting PQ in
BK; also pass a plane through FB and FE, intersecting
PQ in EL,
Then EK \\ FC, &nd BLW FB. Art. 538.
But FE 1 EK and EL. Ait. 505,
.-. FE 1 FC ;.iid FB. Art. 123.
.-. FE ± MX. Art. 509.
.-. MN WPQ. Art, ,^30.
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d'ii BOOK VI. SOLID GEOJIEiny
Proposition XII. Theoresi
535. A straight line perpendicular to one of two parallel
planes is perpendicular to the other also.
\
--
L_4 - \j
\.
.^••J
-1 \
Given the plane MN || plane PQ, and AB L FQ.
To prove AB L MN.
Proof. Througrb AB pass a plane intersecting PQ and
MN in the lines BC and AF, respectively; also throngh
AB pass another plane intersecting PQ and MN in BJ) and
AH, respectively.
Then BC 11 AF. and BD \\ AS.
But AB X BC and BB.
:. AB 1 ^J'and^fl'.
.-. AB X plane J/A".
586. COK. 1. Through a given point to pass a plane
parallel to a given plane.
Let the pupil supply the construction.
537. Cor. 2, Through a ginen point but one plane can
be passed parallel to a given plane.
Art
531.
Art
505.
Art
123.
Art
5CB.
0. S.
>.
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LIMES AND PLANEiS 335
Proposition XIII. Theorem
588. If two angles not m the same plane have their
corresponding sides parallel and extending in the same direc'
Hon, tht angles are equal and their planes are parallel.
Y
Given the Z.BAC m the plane MX, and the ZB'A'G'
in the plane PQ; AB and A'B' \\ iind csteuding in the
same direction; and AC and A'C W and extending in the
same direction.
To prove Z^^iO = Z B'A'C, and plane ilfiV |1 plane P^.
Proof. Take AB = A'B\ and AC=A'C'.
Draw AA', BB', CC, BC, B'C.
Then ABB' A' is a ZI7 , Art. 160,
(foi-ABandA'B'are^and [\).
:. BB' and AA' are = and ll. Art, 155.
In like manner CG' and AA' are = and !|.
.-. BB' and CC are^and ||. (Why ?)
.-. BGC'B' is a C7 , and BC=B'C'. {Why V)
.-. A ABC= A A'B'C. (Why 5)
.-. ^A = ZA', (Why?)
Also AB II A'B', ,■. AB \\ plane PQ. Art. 527.
Similarly AC 11 plane P(?.
.-. plane MN || plane PQ. Art. 534.
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doO BOOK V!, HOLID GEOMKTKY
Propositiux XIV. Theorem
539. // two straight Ihies are hilers€cletJ bf/ tlin^r purah
lei planes, the corresyoiuVuuj svijairnia »/ tlic-'in lines are
proportional.
\
Given the straight iities AB and CD intersected by the
1! planes MN, PQ and SS in the points A, F, B, and 0.
H, D, respectively.
AF CR
To prove —
: the plane PQ in 0.
FB MB
Proof. Draw the line AB Interseetiu
Draw i^G. BB, GH, AO.
Then FG \\ BB, B.\id. GR\\ AG. Art, 531,
.■^'=^. Art. 317.
FB GD
And ^=f|. (W.,n
•■ FB SB ' y'l
Q. E. B.
Ex. 1. la above figure, if AF=1, FB = 5, and CH^S, lind CD.
Ex 2- If Cif=3, SD=4, and JB=10, find JJ-' and fif.
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DIHEDRAL ANGLES
DIHEDRAL ANGLES
540. A dihedral aagie is the opening between two in-
tersecting planes.
Prom certain points of view, a dilitdriii angle may be regarded as
a wedge or slice of space cut out by the planes forming the dihedral
angle.
541. The faces of a dihedral angle aro the planes form-
ing the diliedral angle.
The edge of a dihedral angle is the straight line in which
the faces intersect. p
542. Naming dihedral an-
gles. A dihedral angle may bo
named, or denoted, by naming
its edge, as the dihedral angle
-1J5; or by naming four points,
two on the edge and one on
each face, those on the edge
coming between the points on
the faces, as F^AB-Q. The
latter method is necessary in naming two or more dihedral
angles which have a common edge.
543. Equal dihedral angles are diliedral angles which
<ian be made to coincide,
644, Adjacent dihedral angles are dihedral angles liav-
ing the same edge and a face between them in common.
545. Vertical dihedral angles are two dihedral angles
having the same edge, and the faces of one the prolonga-
tions of the faces in the other.
548. A right dihedj'al angle is one of two c((iial adja-
cent dihedral iuiglcs t'onued by two ])lanea.
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338
vy.
547. A plane per-
pendicular to a givoii
plane is a plaiii; fovra- ^____„
ing: a right diliedra! \
angle with the givuu \
plane . \
Many of the properties of dihedral angles are obtained
most conveniently by using a plane angle to represent the
dihedral angle.
548. The plane angle of a dihedral an-
gle is the angle formed by two lines drawn
one in each face, perpendicular to the edge
at the same point.
Thus, in the dihedral angle C-AB-F,
if FQ is a line in the face AD perpendicu-
lar to the edge AB at P, and PR is a line
in face AF perpendicular to the edge AB
at F, the angle QPIi is tiie plane angle of the dihedral
angle C-AB-F.
549. Property of plane angles of a dihedral angle.
The magtiltmU of tlio plane angle of a dihedral angle is
the same at every point of the edge. For let EAG he the
plane I of the dihedral I E-AB-I) at the point A.
Then PR || AE, and PQ || AC (Art. ll^l.)
.-. IRl'Q - Z.EAC (Art. 538).
550. The projection of a point upon a plane is the foot of
a perpendicular drawn from tlie point to the plane.
551. The projectioa of a line
upon a plane is the locus of the pro- m
jections of all the points of the line
on the plane. Thus A'B' is the
projection oi AB aa. the plane 31N.
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DIHF.DIUL AXGLES 60)}
Proposition XV. Tiieoreji
552. Tico dihedral angles are equal if their plane angles
are equal.
lU .LI I
Given Z DBF t! . | Iiik- I of the dihedral Z ChAB-F,
IB'B'F' the plaiiL. ^ uf the dihedral ^C'-A'B'-F, aud
ZDBF==Z.I)'B'F'.
To prove Z C-AB-F = I C'-A'Ii'-F.
Proof. Apply the dihedral IC'-A'B'-F to IC-AB-F
so that Z li'Ii'F coincides with its equal, IDBF,
Gaom. As. 2.
Then line A'B' mirst coincide with ,4.ii, Art. 515.
{fof A'B' and AB are both X jAu.eiJBFut ihepoiiiIJi).
Hence the plane A'B'D' will coiueido with plane ABD,
Art. 501.
(ihroiigh tao taUmecting lines uiih/ one pliiiie mit he [tti^seil).
Also the plane A'H'P will coincide with the plane ABF,
.". Z (7'-A'/?"-7'"" coincides with Z 6'--4/>'-i'' and is equal
to it. Avt. tr.
558. Cor. The vertical dihedral oni/I''^ Jhnu'-d h 11 tin)
intersecting planes are equal.
In like manner, many otlicr [n'opertii^i^ of phuie aiitrlis
are true of dihedral angles.
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340 )!noK VI. SOLID 'ieoi]i:jrv
Proposition XYL Theorkm
554. Tu-o dihedral aiuih^ li'iv' thr .s^imc r
.■-■,^B
Given Uie Jihe.lral A V-All-D rikI C'-AT,'-!)' liaviug
the plane A fMZ) and C'A'T*', respectively.
To prove I C'-A'B'-n' -. lC-Ali-1) =IC'A'!Y -.
I CAB.
Case I. When the plane A <"A'J>' ami CAT) (Pigs. 2
andl), are comtneiisurahk'.
Proof. Find a eommou measure of tlie A V'A'D' aiid
CAD, as Z GAK, and let it be contained in Z G'A'D' n
times, and in LGAT) m times.
Then / C'A'iy : I CAD^n -. m.
Through A'B' and the lines of division oE Z C'A'D' pass
planes, and through Ali and the lines of division of
Z CAD pass planes. These planes will divide the dihedral
IC'-A'B>-D' into n, and ZC-AIi-D into m parts, all
eqnal. Art, 5S2.
.-. ZC'-A'B'-D' : lC-AB-I)=n -. y,i.
Hence Z C'-xVB'-D' : Z C-AB-7J = Z C'A'D' -. Z CAD.
(Whj ?)
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DlHF-DliAL ANGLES J41
Case II. When (he plane angles C'A'D' and CAB (Figs.
3 and 1) are incommensurable.
Proof. Divide the -Z CAD into any miraber of equal
parts, and apply one of these parts to the I C'A'D' . It
will be eontiiined a i;ertain nnmber of times with a remain-
der, as ^LA'D', less than the unit of measnre.
Hence the i G'A'L and GAD are commensurable.
.-. I C- A'B' ~r,:l C-AB-D^ / fJ'A'L : Z CAD. Case I-
If now we let tho unit of measure be iudefiiiitulv iliniiii-
iahed, the Zi.-l'/)', which is less tliau the unit of measure,
wiil be indefinitely diminished.
.-. I CA'L i I C'A'D' as a limit, and
iC'-A'ir-L^ lC'-A'B'~D' ■i^s-^WmM. An. 2&L
-A'B'-L
Hence
Z C'A'B'-D'
IC-AB-D
becomes a variable,
vith
limit.
But the \
tnes a variable with
^C'-A'B'- L _
Z.C- Mi-B
I C'-A'B'-B'
Ex. I. How many straifjiit lines nre neeessafy to indicate a dihe-
dral nngle (as IE-AB~II, p. 338)? IIow many straight lines are
neeesBary to indieate the plaoe angle of a dihedral angle 1 Hence,
■what is the advantage of naing a plane angle of a dihedral aogle
instead of the dihedral anf;le itfielE f
Ex 2,
ogoug to p:
B thre
iial propert;
[ dihedral angles anal-
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Proposition XVII. Theorem
555. If a tstrhUjkt line is pn-pendic\dar to a plane, every
plane ilmini fliroiigh iluil line in pcrpemUcular to the plane.
Glvea the line Ali X plane MN, and the plane PQ
through A B and interseethig MN in EQ.
To prove PQ X MX.
Proof. Ill the plane .'l/.Vdi-aw BO ± RQ at B.
Bnt AB X JiQ. Art. 505.
.-. ^ABC U the plane / of the dihedral ZP-RQ-M.
Art. 548.
But /.ABC is. a visht I , Art. 505.
(for All 1 MNhy li-ip.).
556. Cor. A plane perpendicular f" the ed<je of a
dihedral angle is perpendicular to eacli of the two faces form-
ing the dihedral aixfjle.
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dihedral anglbs 343
Proposition XVIII. Theorem
557. If two planes are perpendicular to each oilier, a
straight line drawn in one of them perpendicular to their
line of intersection is perpendicular to Ike other plane.
Given the plane PQ X plane 3fy and intersecting it in
the line RQ; and AB a line in PQ X HQ.
To prove AB X plane MN.
Proof. In the plane US' draw BC X EQ.
:. ZABCis the plane I of the dihedral ZP-RQ-M.
Art. 548.
.-. Z^ECisart. /. . Art. 554.
{/hi- I'-EQ-Mu a ri'jht <lihe<lna I ).
.■, AB X BO and fi(^ at their intersection.
.-. AB X plane WN. (Why %)
Q. £. 9.
558, Cor. 1, If two planes are perpendictiiar to each
other, a perpendicular to one of them at any point of their
intersection will lie in the other plane.
For, in the above figure, a X erected at tlie point B in
the plane MX must coincide with AB lying In the plane
PQ and X MN, for at a given point in a plane only one X
can 1)6 drawn to that plane (Art. Slij).
559. CcR. 2. If two planes are perpendicular to each
other, a perpendicular to one "plane, from a point in the other
plane, will lie in the ollu-r plane.
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DOOK Vf. ^OLID GEOMl^TRY
PKOPOSITION XIX, THEOJiEM
560. // iu'O inUrsecHntf plni
a third plane, iheh- Unr of ■in
the third plane.
', each pBrppnd'cnUir
ion is iKvpi-'udicKtar
Given the planes PQ and BS 1 plane Mlf, and inter-
secting in the line AB.
To prove AB ± plane MX.
Proof. At the point B in wliich the three planes meet
erect a ± to the plane MX. This ± must He m the plane PQ.
and also in the plane RS. ^'^^- ^'^'^■
Hence this L must coincide with AB, the iiitersfction
of PQ and RS. Art. 50S, 2.
.-. AB ± plane .1/-V.
g. E. B.
561. Cos. If two planen, indmVnig a right dihfdra!
angle, are each perpendicular to a third plane, the inferspc-
tion of any two of the planes is perpendicular to the third
plane, and each of the three lines of intersection is perpen-
dicular to the other lu'o.
Ex. 1. X»me all iha dihaiiral an^
Ez. 2. If Z CBQ^SO", find the ri
a on the above fijjiire.
o£ ea.eli pair of diliedral d.
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DIHEDRAL ANGLES
Proposition XX. Theorem
562. Every point in the plane which iiseets a given
dihedral angle is equidistant from the faces of the dikfdral
angU-
Given plane CB bisectins? tlie diluidi-al I A-BR-T), P
any point in phme ISC, I'Q iiud FT 1. faces BA and IW,
respectively.
To prove I'Q^PT.
Proof. Through PQ and FT pass a plane interseetmg
AB in QE, BD in KT.and BG in Pli.
Then plane FQT J. planes AB and BD. Art. 555.
.■, plane PQT ± line RB, the intersection of the planes
AB and BD. Art. 5fi0.
.-. RB L RQ, EP and BT. Art. r.or..
.-. i QBP and P/C7' ni^e the plane A of tht dihedral
A A-BR-P and F-BE-D.
But these dihedi-al ii are orinal.
.-. LQRV= IPRT.
:. rt.A P()/^ = rt, A PRT.
:. PQ^FT.
Art. 54S.
Hyp.
563. n^ tonisof all points
of a (lilu'dral angle is the plane t
(Why?)
Q. E. ».
■qni<li.slant from- Ihe fares
^r:ciiitg ihv ilihi'iirul angle.
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1U10K VI. SOLEII
XXI. PiiOBLGM
564. Through unij Ktrdi'jht I'li". not perpendicular to <
giten plane, to pass a plane pcriKiidkular to the given plane
Given the line AB not ± plane MN.
To construct a plane passing through AE and ± MN.
Construction. From a point -1 In ths liue AB draw a
AC to the plane MN.
Throngh the intersecting lines AB and AG
plane AD.
Then AB is the plane required.
Proof. The plane AD passes throngh AB.
Also plane AD ± plane MN,
{M
sAC, ichichig 1 MN).
Art
51(1.
pass
tlie
Art
503.
Conefr.
Art
555.
Q. E.
r.
565. Cor. 1. Through a struigJd line not perpendicular
to a given plane only one plane can be pmsed perpendicu-
lar to thai platie.
For, if two planes could be passed through AB ± plane
My, this intersection AB would be X MN (Art. 560),
which is contrary to the hypothesis.
566. COE. 2. The projection upon a plane of a straigM
line not perpendicular to-that plane is a straight line.
For, if a plane be passed through the given line X to
the given plane, the foot of a X from any point in the
line to the given plane will be in the intersection of the
two planes (Art. 5oy),
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DIHEDRAL ANGLES 347
PROPOsiTiON XXII. Theorem
587. The acute ant/le which a line maki's ivith Us pro-
jeetion on a plane is the least angle which it makes with aiiy
line of the plane through its foot.
Given line AB meeting tlie pianp MN in the point B,
■BC the projection of AB on MN, and PJi any other line
ia the plane MN through B.
To prove that ZIBC is less than ZAIiP.
Proof. Lay off PB equal to CB, and draw AG and AP.
Then, in the A ABC and ABP,
AB=AB.
BG^ BF.
Ent AG < AP.
.: Z^BC is less than lABP
(Why t)
Art. 108.
Q. E.D.
568. Def. The inclination of a ihie to a plane is the
'ciite angle which the tjiven line makes with its pi'ojsetion
'Pon the given jjlaiie.
E». 1. A plane has an inelination of 47" to ea«tt o£ the ta^ne of a
iliadral angle and is parallel to clie edge of tbe dilieiiral angle; how
aonj degrees are in the pkne UHglt of the dihedral an^'l..';
Ex. 2. In llie fiL-iire ™ pae? 3-ir,. if PT=QT,liow large lathe
'''"dral z A -!:il-li'i i[ n---llT, hmv large Is it?
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SOT.ID gi:u.mk-i'i;y
Proposition XXIII. Problem
569. To <lruw a common pTpeu'/'Cidar to any tivo Hi,
tiof in the same plane.
Given the lines AB and VJJ not in the same plane.'
To construct a line perpenLliculiir to botli AB and CD.
Construction. Tlirongh AB i>i(..3=! a. plane MXW line CIk
Art. j>!
Through CD pi^s a piano CF 1 ])laiie M^' (Art. 564), aui
interaecting plane MN in the line BF.
Then EF \\ CD (Art. 528), .-. EF must intersect Ai
(which is not |1 CD by hyp.) in some point K.
At E in the plane CF draw LK ± EF. Art. 274
Then LK is the perpendicular reqitired.
Proof. LK ± EF. Con^ir
.-. LK ± CD. Art. l:.;.
Also LK ± piano MX. Art. ss:
.-. LK _L line _l7i. (Wl,y^
.-. LK ± both C« and AB.
Q. E. ?.
570. Only one perpendicular can be draicn between ttco
lines not in the same plane.
For, if possible, in the above figure let another Hue BD
be drawn ± AB and CD. Then, if a line be drawn throagh
B II CD, BD X this line (Art. t2;t), and .-. X plane MN
(Art. 509). Draw DF X Hue EF; then Z>F X plane MS
(Art. 557), Henoe from the point T) two X, DB and DF,
are drawn to the plane MX, which is imposBiblu (Art. 517),
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rOLYIIEDUAL Af^dLK
POLYHEDRAL ANGLES
571. A polyhedral angle is the amount
o£ opening between tlireu or more planes
meeting at a point.
Such an angle may be regarded as a portion of
spBoe cut out by the planes torming the anfila.
572. The vertex of a polj-hedral angle
is the point in which tlie planes forming the angle meet ; the
edges are the lines in which the pianes intersect; the faces
are the portions of the planes forming the polj"lie(ival angle
which are inclnJed between tlie edges ; tlie face angles are
the angles Eormeil by the edget;.
Each two adjacent fae.es of a polyliedral angle form a
(iihedral angle.
The parts of a polyhedral angle ;ire its faee angles ami
liihedval angles taken togetlier.
573. Naming a polyhedral angle. A polyhedral angle
ia named eitlier by namiuij the vertex, as I': or by naming
the vertex and a point on each edge, as V-ABG.
In case two or more polyhedral angles have the same
Vertex, the latter method is necessary.
In the above polyhedral angle, the vertex is Y; the
edges are VA, YB. T"C; the face angles
are J.rjS, BVC, AVC.
574. A convex polyhedral angle is ii
polyhedral angle in which a suction
ttiade by a plane cutting all the edges
IS a convex polygon, as y-ABCI>E .
575. A trihedral angle is a polyhe-
iral angle having tliree t'aces; a tetrahedral angle is one
laving four face*, etc.
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GEOJIETliY
576. A trihedral angle is rectangular, birectangular, or
trirectaogular, aeeordinK as it contains one, two, or /ht-ee
right dihedral angles.
577. An isosceles trihedral angle i
two of whose faee angles are equal.
trihcdi-al angle
578. Vertical polyhedral angles an
having the same vertex and the faees (
the other produced,
579. Two equal polyhe-
dral angles are polyhedral an-
gies bavin;^ their correspond-
ing parts equal and arranged
in the same order, as V-ABC
and V~A'B'C.
Two equal polyhedral angles
may be made to coincide.
580. Two symmetrical poly-
hedral angles are polyhedral an-
gles having their corresponding
parts equal but arranged in
reverse order.
If the faces of & trihedral angle, V-ABC,
produced, they will form a vertical trihedral angle,
V-A'B'C, which is symmetrical to V-ABC. For,
if V-A'B'C be rotated forward about a horir.ontal
aiis through V, the two trihedral angles ai
to have their corresponding parts equal I
ranged in reverse order.
Similarly, any two verlieal polyhedral
are aymmetrieai.
polyhedral angles
: one the faces of
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P0I.YI!E1>R.\L ANGLES
351
581. Equivalence of symmetrical polyhedral angles. It
has been shown in Plane Geometiy (Art. 488) that two
triangles (or polygons) symmetrical with respect to an
axis have their corresponding parts equal and arranged
in reverse order. By sliding
two such figures about in a plane
they cannot be made to coincide,
but by lifting one of them up
from the plane- in which it lies and tnrnii
be made to coincide with the other figure.
Symmetrical polyhedral angles, however, cannot be
made to coincide in any way; henee some indirect method
of showing their equivalence is necessary. See Ex. 29,
p. 358. and Arts. 789-792.
it mav
8 figUl
Prop. XX. rf
those OQ tlie figure !
Kr, 1. Name the trihedral angles
i^PBQ^m", what kind of trihedral a
If iLPBQ=m°, whot kind are they?
Ei. 2. Are two trireot angular trih
Prove this.
Ex. 3. Are two lines which are perpendicular to the same plane
necesaarily parallel f Are two planes which are pei'pendicular to the
same plane necessarily parallel? Are two planes which are perpen-
dicular to the same line necessarily parallel f
Ek. 4. Let tlie pupil out out three pieces of pasteboard of the form
ndioated in the aoeoropauying figures; cut them half through where
'ho lines are dotted; fold them and fasten the edges so as to form
hree trihedral angles, two of which (Figs. 1 and 2) shall be equal
iEd two (Figs. 1 ;ind 3) sjmmetrical. hy esperiment, let the pupil
ind which pair may be made to ooinuide, aud whifh not.
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V[. SOLID GEOMIlTILV
I'KOl'OfcJTIOX SXIV. TUEOKEM
582. Thr sum of nnij fro f>w<' mujU, of a trlhedru!
an'jtc is greater than the third faca atKjh.
Given the trihedra! angle. S-ABC, with angle A8C its
greatest face angle.
To prove lASB + I HSC gvaai-n' {linn lASC.
Proof. Ill the faee AS(^ ihviw n;>, itiakiiig ZASJ)^
ZASli.
Take *7>=,S7>'.
In the faee ASC draw the Hne AUC in any convenient
direction, and draw AB and BV.
Then, in the A A8B and ASB, SA^SA.
SB = SD, and Z,l«fi= ZAS'I).
:. A A8B=A ASVf.
.-. AB=AD.
Also AB + BC> AC.
Hence, subtracting the equals AB and AJ).
BC> BC.
Hence, in the A BSC and DSC, 80=^ 8C, «B=Si>, and
BC > DC. (Why!)
.-. /. BSC is greater than Z DSC. Art. 108.
To each of these uneqnals add the equals ZASB and
ZASD.
:. ZASB+ :/ B«0 is greater than I ASC. (Whyt)
0. E- D.
Ex. Iq the above figure, if IAj'<C equals one of the oilier £ace
angles at S, as lASB, bow is the theorem proYod J
(Why!)
(Wl,y=)
(my ?)
(Why?)
(\Vby?)
(Why!)
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POLYHEDKAL ANGLES o5d
Proposition XSV. Theorem
588. The sum of the face angles of any convex polyhedral
angle is less than four right angles.
Given i.l!c polyhedral angle S-ABCDE.
To prove the sura of the face ^ at 5 less than 4 rt. ^ .
Proof. Pass a plane cutting the edges of the given poly-
hedral angle in the points A, B, C, J), E.
Prom any point in the polygon ABODE draw OA.
OB, OC, oh, OE.
Denote the A having the common vertex S as the S A,
atid those having the common vertex O as the A.
Then the sum of the A of the S A=^ the sum of d of
the A. AH, 11)4.
Bat ZSR'l+/S7J0iagi-eatert!ian lABC, 1 , . „„
Z5C/J+ ZSCiMsgreaterthan ^BOT, etc.) '
•'. the sum of the base <i of the S A > the snra of the
tase A of the A. Ajt, 9.
.■. the sum of the vertex d of the S A < the sum of
Uie vertex A of the A, As. ii.
{if unequal s be suhtratilid from ctiiiiih, Ihe remainders are iiiicqiinl
iiirermeor'Ui).
But the sum of the d at 0=4 li. i . (Wiiy?)
.'. the sum of face d at ii < 4 I't. A . Ax. s.
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<i04 BOOK V!. SOLID GEOMKTRY
PnOPOSITlOX XXVl. TlUXlREM
584. JS iii-0 Irihedml ainjhs luu-c ihe thnc face migh^,
of one eqmtl to the tliree fuce uiigl'-s of ilie other, ike (yi-
hedral angles have their correspiinding dilwdml niifjUsequn}.
and are- edlher equal or sijmmptriruJ . iircordiiir/ as fl/d,-
corresponding face angles are arraniU'd in, f/w mine or Ui
reverse order.
Given thf trihedral i ^-ABC and ^'-A'B'C, liavin-
the face A AUB, ASO 3.\\<\ BWC'equal to the face A A'ii'li'.
A'S'C and B'S'C, respectively.
To prove that the con-espon<3ing dihedral A of S~ABf'
and S'-A'B'C are equal, and that A S-ABO and N'-
A'B'C are either equal or symmetrical.
Proof. On the edges of the trihedral A take ^S'.l, .S7>',
8C, S'A', 8'B', S'C ail equal.
Draw AB, AC, EC, A'B', A'C, B'C.
Then, 1, In the A ASB and A'S'B', SA = S'A', SB =
S'B', and AASB= ZA'S'B'. (Why?)
.-. A ASB^A A'S'B'. (Why!)
.-. AB^A'B'. (Why?)
2. la like manner AO=A'C', and HC = B'C".
:. A ABC^A A'B'C. fWhyf)
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EXERCISES OS THE LINE AND PLAKE 355
3. Take D a convenient point in SA, and draw DE in
the face ASB, and DF in the face ASC, each ± SA.
DE and i>F meet AB and .10 in points E and JP,
respeetivelv.
Similarly, take S'lf^SD and construct A J>'l-yr'.
Then, in the rt. ^ ADE and A'B'E', Alt^A'l)', and
Zi>jli7-./rA'l,". (Why?)
.-. A ;inj;-A A'D'E'. (Why?)
.-. A/^'=^'i", !ind !)E=I)'E'. (WliyM
4. In like manner it m:\y\K sho^yu that AF=A'F', and
.-. A .4Z:F=A J.'£'F'. (Wii;-?)
And EF^E'F' (Whyf)
5. Hence, in the A Z)i;F and B'E'F', T>E=n'E', I>F
= D'F and EF=E'F'. (Why ?)
.-. A DEF^A D'E'F'. (Why f)
.-. IEJ}F=IF:J)'F. (Why?)
Bat these i are the plane i of the dilieitral A whoso
edges are S.l and N'A'.
.-.dihedral Zif-Afr-C^diliedrnl IB'-A'S'-C Art. 5.12.
In like manner it may be shown that the dihedral A at
SB and S'B' are equal; and that those at KO and S'C are
equal,
.'. the trihedral A S and S' are either equal or sym-
metrical. ^^'^^- 5'9' ''^o.
Q. E. B.
EXERCISES. CROUP 6't
THEOREMS CONCEKNIXG TUE LINE A\D f'LANE IN SPACE
Ex. 1. A segment of a liiie not parallel to a plane is longer thira
its projection in the plane,
Ex. 2. Equal stE-aight lines drawn from a point to a place are
equally incliiit-a to the plane.
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35G B
Ex. 3. A line
parallel.
Ex. 4. If thre
perpendicular to a
SOt-in GEOMETRY
H perpeudiculiir to the si
ig in three stniight lines acs
£ iulerscction Jii't parallel.
Ejt. B. If a plane bisects any line at riglit
angles, any point in the plane is equidistant
from the ends of the line.
Ex. 6. Given JIl X plane .If-V,
and ,)(.' ± piano SS;
prove JIC ± yii.
Ex. 7. Given PQ X plane .1/-V,
PR X plane HL,
and BS 1 plane J/A';
prove QS X AB.
Ex. 8. If a line is perpendicular to one
of two intersecting planes,. its pro.jpctioQ
on the other plane is perpeniJiQuliir to the
line of intersection of the two planes.
Ex. 9. Given CE X BE,
AE I T>E,
and Z C-AD-E a rt. dihedral Z
prove CA X plane DAE.
Ex. 10. The projections of two parallp
a plane are parallel. (Is the converse
theorem also true ?)
Ex. 11. If two parallel planes are cut hy two non -parallel planes,
the two lines of intersection in each of the parallel planes will make
equal angles.
Ex. 12. It a line is perpendicular to a plane, any plane parallel
to the line is perpendicular to the plane. (Is the converse true !)
Ex. 13. In the figi
DC; prove BF X DC.
Prop. VI, given AFi X MN and AF X
X DC.
Ex. 14. Two planes parallel to a third plane are parallel to each
[Sua. Draw a line X third piano.]
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EXERCISES ON THE LIKE AND PLASB
Ei. 15.
straight liii
The
projoetioiis up
9 equal nnd p^r
on a plane of t
illel.
Ex. 16.
section.
Al
ne parallel to t
vo pluues is p
Ki. 17.
prove the a
la
ngle
the flguro to Py.p. XXII. it
JRi' obtuse.
Ei. 18. In a quadrilateral in apace (i. e., a
quadrilateral whose vertices are not all in the
same plane), show that the lines .ioliiing thu
midpoints of the sides form u parallelogram.
Ex. 19,
the oppoait
The
esid
lines joiniug the midpoints o
s of a quadrilateral in Space Ij:
Ex. 20.
three faces
The
planes biseeti
Hue every po
g the ilihedral
ut of wliii;h is
[SUQ,
Sbo.
rt. m2.]
Ex. 21
7
n 0(J bisecting I ROS,
PQ X plane EOS,
QR X OR,
QS X OS;
vo PIl=PS, PR X OR,
□ii PS X OS.
■qual and par
pavallel to their ii
Ex, 22. In n plane bisecting a piveu plane nn^^e, and perpendicu-
lar to its plane, every point ia equidistant from the sides of the angle.
[Sl'q. See Ei. 21 ; or through P any point in the bisecting plane
pass pianes X to the aides of the iL , etc.]
Ex. 23. In a trihedral angle, the three planea bisecting the three
face angles at right angles to their respective planea, interaeet iu a line
every point of which is equidistant from the three edges of the tri-
hedral angle.
Ex. 24. If two faceangleaof a trihedral augle are equal, the dihe-
dral angles opposite them are equal.
Ex, 25, In the figure to Prop, XXIV, provi) that lASC-'r IBSC
ia greater than Z ASl) -{- Z ISSU.
Ex. 26. The common perpendicular to two lines in spui^e is tho
shortest line between thero.
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'■^A
vwr\
i;noi; vt, rood
. Given .l/.Vll Jin,
and A AIIC => A iibc.
trihedral angles are equal.
El. 29, Any two sjmmelrieiii trih,
Aval angles are equivalent.
[Suo. Take S.l, Sli, sr, S'a'.
S'li', S'C', all fqiml. Pass planes -IfiC,
A'B'C. Draw SO aiid S'ty ± tlieso
plitnes. Tben thw trihedral A am
divided into three pairs of isoseelos
symtuetripal trihedral d , etc.]
EXERCISES. CROUP 68
LOCI IN SPACE
Find the loeiis ol a point equidistant from
Ex. 1. Two parallel planes. Ex. 3. Three t;ivea points.
Ex. 2. Two given points. Ex. 4. Two interseoting linei
Ex, 5. The three faces of a trihedral angle,
Ex. 6. The three edges of a trihedral angle.
Find the ioeus
Ex. 7. Of all linos passing tlirongh a si^en point and parallel
L plven plaice.
Ex. 8, Of all liuPB pcrptiKiicul
ar to a
given line
It a given -point
n the line.
Ex. 9. Of all points in a give
n plani
3 eqnidisti
mt from a given
loint outside the plane.
Ex. 10. Of all points equidista
ut froii
1 two £ivei
1 points and from
,wo parallel pianos.
Ex. 11. Of all points equidista
lit frou
1 two give]
11 points and from
Swo intersecting planes.
Es, 12. Of all poinis at a give
■n dista
iicu from
a given plane aod
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EXERCISES OK THE LINE AND PLANE 6Di}
EXERCISES. CROUP 66
riiOllLKMS CONCERNING THE POINT, LINE AND PLANE
IN SPACE
Ex. 1. Through a given point pass a plane parallel to a given
plane.
Ex. 2, Through a given point [lass a plane perpsniiii:iilai' to a
given plane.
Ex, 3. Tlirougii a given point to (construct a plani' paralli'l to two
given lineR which are nol in the suiiie plai^f.
I'rovf that only oui! piaue oun be coiistrurli^il fiilfimii!; Iho givi^n
conditions,
Ex. 4, Bisect a given iliheilral angle.
Ex. 5. Draw a plane equallj inclined to three lines whleh meet at
a point.
Ex, 6. Through a piven point draw a line parallel to two givan
intersecting planes.
Ex. 7. Find a point in a plane .such that lines drawn to it from
two given points without the plane make equal angles with the plane,
[SuQ, See Ex. 23, p. 176.]
Ex, 8, Find a point in a given line equidistant from two given
points.
Ex, 9. Fhui a point in a plane equidistant from three given points.
Ex. 10, Find a point equidistant from four given points not in a
plane,
Ex. 11. Through a given point draw a line which shall intersect
[Sua. Pass a, plana through the given point and one of the given
lines, and pass another plane through the given point and the other
given line, etc.]
Ex. 12, Through a given point pass a piano cutting the edges of
a tetrahedral angle so that tlie section shall be a parallelogram.
fSVG. Produce each pair of opposite faces to interseet in a
Straight line, etc.]
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Book VII
rOLYL{EDRONS
585. A polyhedron is i ^lolid buiimlnl \i\ pUnes
586. Tha faces o£ a pohlie
6vou ar(i itri houmliiiy: pfiiiiLS tin,
edges of a polyliedrun are tlii. hne% i
of intersect ion of its faces
A diagonal of a polyhedion lo i
a straight line joining two ot ita
vertices whicli are not m the j
same face. The vertices of i polv-
hedrou are the points in wliiji iti,
edges meet or intersect.
587. A convex polyhedron is a i>olylK'dvo!i in which a,
section made hy any phme is a convex polygon.
Only convex polyhedrons are to be considered in Ibis book.
588. Classification of polyhedrons. Polyliedrons are
simetimi'i elibsififd tttoidm^ to th mimlei of their
fices Thus a tetrahedron it, a pohhedion ot foui fices
T hexahedron is a pohh^dion of -ii^ fn. ^ in octahedron
1 n rf tiglt I dodecahedron one if tftcl\c ml an
cosabedron one ot twcntj f^ces
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POLYHEDKOXS
The polyhedronB moat important in practiea! life are those deter-
mined by their stability, the facility with which thijy can be made out
of eommoQ raateviuls, a3 wood and iron, tha readiiieaa with which
they can be packed togethti', etc, Thus, pfism meaua "aomething
Bawed oil."
PEISMS AHD PARALLELOPIPBDS
589. A prism is a polyhedron bounded
bj two parullcl planes acd a group of
planes whose Hues of iiitersectiou ui ""
parallel.
590. Tlie bases of a prism are tbe
faces formed by tbe two parallel planet.,
tbe lateral faces are the faces formed h-\
the group of planes whose lines of inteiseetKm aie paiallel.
The altitude of a prism is the perpendicular distance
between the planes of its bases.
The lateral area of a pi-ism is the sum of the areas of the
lateral faces.
591. Properties of a prism inferred immediately.
1. The. lateral edges of a prism are equal, for they are
parallel lines iucluded between parallel planes (Art. 589)
and are therefore equal (Ai-t. 532).
2. The lateral faces of a prism are parallelograms (Art.
160), for their sides formed by the lateral edges are equal
and parallel.
3. The bases of a prism are equal polygons, for their
homologous sides are equal and parallel, each to each,
(being opposite sides of a parallelogram), and their liomoi-
o?ous angles are equal (Art. 538).
592. A right section of a
plane perpendicular to thu hit'
priM
section made by a
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3li2 1;00K VII. SOLID ceometisy
C.O3 \ trianiFiilnr nri«m \< n iirUin ivlio';.' liilKf
n I
it
594. All oblique prism is ii prism wliose Literal edges
i-e ol)lir|iie to the liases.
595. A right prism is a prism whose
lateral {idgcs are pei'pemlieiitar to the bases.
A regular prism is a right prism f
ses are regiilai- polygons. 1
whose bases
597. A truncated prism is that part <if
a pvisiii included between a base and a
section nnide by a plane obliiine tu llie
base and cutting all tlie lateral edges.
598 \ parallelo piped j wl e
ii II n r I'll ' " '
599. A right parallelepiped is a purallelopiped whose
lateral edges are perpendicular to tiie bases,
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PmSJtS AND rAKALLELOPIPEDS 363
600. A rectangular parallelopiped is a right pavuUelo-
piped whose bases are rectanglos.
Hence, all the fitces of a rcciuiujidur puralldojyipcd are
rectanglns .
601. A cube is a rectangulai- parallelopiped whose edges
are all equal.
Hence, all the faces of a ciihe are s(iHares.
602. Tlio unit of volume is a eube whose edffe is equal
to some linear unit, as a cubic inch, a, cubic foot, etc.
603. The volume of a solid is the uuraber of units of
volume which the solid contains.
Being a iiwHiber, a volume may often be dptermitied from other
numbers in certain expeditious wiiys, wliioii it is one of tlie objects of
geometry to determine.
604. Equivalent solids are solids whose volumes are
equal.
it number of faces whiol
El. 2. A square right prism is what JEind of a parallelopiped f
Ex. 8. Are there more ri^ht parallelepipeds or rectangular paz'al-
lelopipedst That is, which of tliese indiides tUe other as a special
<;aser
Ex. 4. Prove that it a given straight line is perpeinlicular to a
Ten plane, and auotber straiffht linn is perpendicular to another
i-aae, aad the two planes are parallel, Iheu the two given lines ara
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BOOK VII. SOLID GEOilETHY
Proposition I. Tiieoiiem
605. Serliona of a piiam vtadc hij pantUd planes cuUing
all the lateral edges are equal pohjijvns.
Given tlie prism PQ out by || planes forming the sections
AB and A'B'.
To prove section ^7J = secttion A'B'.-
Proof. AB, BC, CI), etc., arc || A'B', B'C, G'ly, etc.,
respectively. Art. 53i.
.-. AB, BC, CI), etc., are equal to A'B', B'C, C'D'^
etc., respectively. Art.isr.
Also A ABC, BCD, etc., nre eqiuil to AA'B'C, B'L'l)',
etc., respectively. Art. 538.
.-. ABCDE^A'B'C'D'E', Art. 47.
Q. E. B.
606. Cor. 1. livery section of a prism made by a plane
parallel to Ike base is equal to the base.
607. Con. 2. All fiylU sneliuns of a pt-ism are equal.
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PRISMS 365
Proposition II. Theorem
608. The lateral area of a prism is equal to the product
of the perimeter of a right section hy a lateral edge.
BX^aCJ GJ^BCXE,
area £17 IQ = CT>XE, etc.
Given the prism KQ, with its lateral area iloiioteil by 8
and lateral edge by E; and AD a right section of the givou
prism with its perimeter denoted by P.
To prove S^PXE.
Proof. In the prism liQ, each lateral edge = /7, AiLTjOI, i.
Also AB ± GH, BG X IJ, etc. Art. 505.
Hence a.vez.CJ RE^AB X GE^AB X ^,1
■ni^' ^ V Art. ?.Sr>.
But 8, the lateral area of the prism, equals thi> sum of
the areas of the ZX7 forming the lateral surface.
.-. adding, 8'^UB + BG+ CV + etc.) X E. Ax. 2.
Or S^PXE.
q. E. D.
609. Cor, The lateral area of a right prism eQv,als the
product of the perimeter of the b<tse hy the altitude.
Ex, Find the lateral area of a rit'lit prtam whose nltitiide U 12 in.,
and whose buso is an equilateral triangle witti a side of G in, Alsa
flad the total area of this lijfuro.
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otlfl BOOK Vn. SOLID GEO^IETItY
PiiorosiTiox III. Theorem
610. If in-o pnf.ms Jiave the lliirp fa-'cs mcludln// a
trihi'iJrid angle of one eqmd, rf^pectireli/, to the three faces
iiidiidhifi a trihedral angle of the oOier, and similarly
placed, the prisms are equal.
Given tlie prisms AJ and A'J', having the faces AS,
AJ>, AG equal to the faces A'K', A'D', A'G', respectively,
suA similai'ly placed.
To prove AJ=A'J'.
Proof. The face d EAF, EAB and BAF arts equal,
respectively, to the face d E'A'F', E'A'B' and B'A'I". Hyp.
.-. trihedral ZA = trihedral AA' Art. 584.
Apply the prism A'J" to the prism AJ, makine each of
the faces of the trihedral /.A' coincide with corresponding
equal face of the trihedral LA. Geom.Ai.2.
.". the plane F'J' ivill coincide in position witJi the
plane FJ. Art. 500.
{/"I- tliepnbils C, P, K' coincide Kith G, F, K, respectively).
Also the point C will coincide with the point C.
,', Cfl'wili take the direction of CH. Geom.Ax. 3.
.'. S' will coincide with H. Art. 508, i.
In like manner J' will coincide with J.
Hence the prisms AJ and A'J' coincide in all points.
.-. AJ=A'J'. Art, 4T.
611. Cor. 1. Two truncated prisms are equal if the
ihrep. faces i)icliiding a trihedral angle of one are equal to
the thr&e faces including a trihedral angle of the other.
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612. COK. 2. Two right prmns are equal if they have
egiMii bases and equal altitudes.
Proposition IV. Theorem
613. An oblique priavi is equivalent to a riqht prism
whose base is a right section of the oblique prism and whose
altitude is equal to a lateral edge of the oblique prism.
Given the 'oblique pvism AD', with the right section JV";
also the rigrht priKin FJ" whose lateral edges are each equal
to a latei-al edge of Alf.
To prove AD' ^FJ'.
Proof. AA' = FF. Hyp.
Subtracting FA' from each of these, AF=A'F'. ("Why ?)
Similarly BG = B'(}'.
Also An = A'B', and FO^FO'. Art. 155.
And A of fare iH7 = homologous A of face A'G'. Art, 130.
.-. face AW^f.iee A'G', Art 47.
(fur ihey haco all tlieir parts eijuul, each to each, and .'. can he made
In like manner face -iK^=face A'K'.
Eat faee Ai) = Eace A'ly. Art. 5B1, 3.
.■. truncated priam j4J=tninc.ated i)rLsm A'J'. Art, oil.
To each of these equals add the solid i-7/.
.-. AB'^FJ'. (Why?)
Q.B. ».
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3G8 EOOK VII. SOLID GEOJIETUY
PfiOTOSITlON V, ThEOEEJI
614. The oppo^Ht Uiferal facfs of a paralli'Iop-ipcd are
equal U)id parallel.
Given the piiralleSo piped AH with tiie base AG.
To prove AG = an.d \] EH, anil AJ= aud || BH.
Proof. The base AC is a /ZJ . Art. 598.
.-. An = am\ !! EC. (Wiiy?)
Also the lateral face AJ is a CH . Art. 591, 2.
.-. ^F^and || EJ. (Whjf)
.-. inAF=ZCEJ. Art. 539.
Aua £Z7AG^/-y EH. Art. m.
Also pUne A(7 11 plane EH. Art. 538.
In like manner it inay he shoivii thitt AJ and HH are
eqnal and parallel.
0. E. D.
615. Con. All// tiro oppoi^ite faces of a pandlelopiped
may be lalrn «n the ba.'ics.
Ex. I. How roBny edKes has a paruHelopiped ! How many faoea ?
How many dihedral angles ? How many trihedral angles f
Ex. 2. Find the lateral area of a prism whose lateral edge is 10
Ex. 3. Find the lateral area of a right prisni whose lateral edge >a
IG and whose base is a rhombus with diagonals of 6 and 8 in.
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Proposition VI. Theorem
616. A plrnie passed throngli liro didgiyii'dhj opposite
edges of a paralklopiped divides the parallelopiped into two
equivalent triangular prisms.
Given the parallelopiped AE with a plane passed through
the diagooally opposite edges AF and CS, forming the tri-
angular prisms ABO-G and ABO-E.
To prove ABC-G^^ADC-K.
Proof. Construct a plane X to one of the edges of the
prism forming the right section PQRS, having the diagonal
PH formed hy the intersection of the plane FHCA.
Then PQ 11 Sn, and QR \\ PH. (WhyT)
.-. PQRS is a ZZ7 . (Why ?}
.-. A PQB= A PSE. (Why ?)
But (lie triangular prism J.BC-0 ~ a prism wlinse hase
is the right section PQR and wliose altitude is AP. Art.6i3.
Also the triangular prism ADG-K o a prism whose base
's the riglit section PSIi and whose altitude is AF. (Whyt)
But the prisms having the eqnal bases, PQE and P8R,
SQd the same altitude, AF, are eqnal. Art, G13.
.-. ABC'G ^ AD(J-K. A3 1.
q. E. ».
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dii) BOOK VII. SOLID Gr.o:^n:TiiY
Proposition VII, ThEOI!EJI
617. If two rectangular parallelopipitiWImvii equal bases
thru "'■^ ^0 each other as their allihules.
Given the rcctaiiguhir parallnlo|Hpi.';lB P' and P having
equal bases and tlie dtitutles A' li' ami AH.
To prove P' -. P^A'B' -. AD.
Case I. When ike altitudes AT/ iniil AP, 'are com-
mensurable.
Proof. Find 'a eoimnon measure of A'li' and AB, aa
AK, and let it lie contained in A'Ji' n tiinos and in A}i m
times.
Then A'B'_: AB^n -. vi.
Tlii'ough the points of division of A'B' and AP> pa>,3
planes parallel to the bases.
These planes will divide I" into ii, and P into m small
reetansjukr parallelopipeds, all erpml. Ai-t. 612.
:. r : P = n: m.
.: F : P==A'B' : AB. {Why ^
Case II. When the altitudes A'B' and AB are incoin-
mcn.'iurable .
Lut the pupil supply the proof, using the method of
limits. (See Art. 554).
618, Def. The dimensions of a rectangular parallele-
piped are the three edges whieh meet at one vertex.
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l'AKAI.LELOFIPEDS
619. Cor. If two rectangular partiUelopipeds haw two
dimensions hi common, they are to each other as their third
dimensions.
PHOPOSITION VIII. TlIEORRM
620. Two rectangular ■jxirallclo'pipcds having equal alti-
tudes are to each other as their bases.
Given the rectangular parallelopipeds P and P' having
the common altitude a, and the dimensions of their hases
6, c and &', c', respectively.
P ^ hXe
P' b'X c'
Proof. C«iii^tri.R't 1 he rectangular parallulopiped, <i>, whose
altitude isit and the dimensions of who^e base are h and c'.
To prove
Then
Moltiplj'ing the corresponding mciubers of thes
equalities,
621. Cob. Two rectangulay paralldopipedH having one
'U'ltiension in common un- to each other as the products of the
other iiro diiiicnsioiis.
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iSiJ BOOK Vn. SOLH) liEOMIiTKY
Proposition IX. Theorem
622. Ai"j Ih-o rpciangulnr pfirnllelopipeds are to each
other as tlw pro'liic/a of ilinr thne dhi
Given tht! rectaiigulai- parailuloplpeds Paiid 2" havir
the dimensions «, h, c and a', V, e', respectively.
To prove
' X fi' X >:'
Proof. Coiistnict the reetaugular panillolo piped Q hav-
ing the dimensions a, li, c'.
P r.
Then 7}^~' Art. cis.
Multiplying the corresponding members of these
equalities,
P_ » X&X c
P a'Xb'Xc'' ^''*'
Q- E. ».
Ex. 1 . Find the ratio of the volumes of two reetaagular paraOelo-
pipeds whose edges are 5, G, 7 in. and 7, 8, 9 in,
Ex. 2. Which will hold mote, a bin 10 s li 1 7 ft., or one Sx
4 s 5 ft. 1
Ex. 3. How miiuy Liicks 8x4x3 in. are neeesaary to build a wall
80s6 ft. sH ja. !
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PAliALLELOPirEDS
Proposition X. Theorem
623. The volume of a rectangular paralleJopiped is eqtial
the product of itn three dimensions.
Given the rectangular para lie loin [jed F having the three
dimensions a, 6, c.
To prove volume of P~a X 6 X c.
Proof, Take as the unit of volame the cube C, whose
edge is a linear unit.
Then - = aXbXc ^
U 1X1X1
The volnoie of P is the number of times P containK the
unit of volume U, or — Art. 603.
.". volume of P=« X 6 X i;.
624. Cor. 1. The volume of a cube is the ciihe of its
edge.
625. Cor. 2. TM vobime of a rectangular paraUelo-
piped is equal to the product of ita base by its altitude.
Ex, 1. Find the numljer of cubie iuchtg in the volume of a cube
whose edge is 1 ft. 3 iu. How many bushels dots this box contain, it
1 bttshel=2150,42ou. in. t
Ex. 2. Tlie meaBurempiit ot tiie volume if a OAibe reduces to the
it of the length of what aiiigiu striiiglit line f
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374 B0(1K VII . SOLIll GEOMETRY
PROPOSITION XI. THBOai:.1I
626. The volume of any pamllelopiped is' equal to th^
product of its base by its altitude.
Given the oblique pai-al!t?lopiped P, with its base denoted
by B, and its altitude by H.
To prove volmne of P=7J X E.
Proof, In P produce the edge CD and all the edges
parallel to CD.
Ou CJ) produced take FQ^CB.
Pass planes through F and G ± the produced edges,
forming the parallelepiped Q, wltli the rectangular base
denoted by B'.
Similariy produce the edge GI and all the edges !1 GL.
Take IK=OI, and pass planes through I and A' X the
edges last prodneed, forming the rectangular parallclopiped
B, with its base denoted by B".
Tlion
P.
>e^
>R.
Art, G13
Also
£<
sP'
-P".
Art. 3S6
But
volume
ofB
.B
XP.
Art. 625
.'. volume
ofP-
.B'
Xff.
Ax. 1
Or volume
[OutliEe Proof. F
OtP
=.« =
-P X U.
S^B" KM.
= P
Xil.]
Ax. 8
«. B. P.
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Proposition XII. Theorem
627. The volume of a triangnlur prism is equal to the
product of Us hu^e by its aUitude,
Given the triangalar prism PQB-M, witli its volume
denoted by 1', area of base by B, and altitude by JI.
To prove Y=BX II.
PQ, Qii. Q^J^, construct the
Proof. ITpou the edg
parallelopiped QK.
Hence (^ff^twice FQH-M.
But volume of QK=in-iin PQBTX R.
= 2B X H.
.: twice volume PQR-M==2B X H.
:. volume PQRSI^ B X H.
Ex. 2 Find the volumo o! a
and tLo edata ol whose base ar
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376 HOOK VII. SOT.TD GK(DMETHY
PROPOSITION XIII. Theorem
628. TJie fnltinie of (inij prism is vqual to the produU of
Us base bi/ its altitude.
Given tlie prism AK, with its vohime dmioted by V,
area of base by B, and altitude by H.
To prove V=BXE.
Proof. Through any lateral edge, as AF, and the dlag-
onaJs of the base, AG and ylJ), drawn from its foot, pass
planes.
These planes will divide the prism iuto triangular
prisms.
Then V, the volume of the prism ,-17i, equals the sum
of the volumes of the triaugnlar prisms. As. G.
But the volume of each triangular prisni = ita base X 11.
Hence the sum of the vohiiues of the A prisms = the
sum of the bases of the A prisms X II.
= liy.U. Ax. 8.
.-. V=BXE. kn. 1.
<}. E, D,
629. Cor. 1. Two prisms are to each other as the pro-
ducts of their bases by their altitudes ; prisms having equiva-
lent bases and equal altitudes are equivalent.
630. Cob. 2. Prisms having equivalent bases are to
each other as their altitudes; prisms having equal altitudes
are to each other as their bases.
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377
PYRAMIDS
681. A pyramid is a poiyhcdrou
bounded by a. group o£ planes passing
through a common point, and by another
plane cutting all the planes of the group.
632. The base of a pyramid is the
face formed by the cutting plane; the '
lateral faces are t!ic faces formed by the
group of planes passing through a com-
mon point; the vertex in the common point through wliii-h
the group of planes passes; the lateral edges are the inter-
sections of the lateral faces.
The altitude of a pyramid is the perpeodicular from the
vertex to the plane of the base.
The lateral area is the sum of the areas of the lateral
faces.
638. Properties of pyramids inferred immediately.
1. The hdn-al fan:^ of a pi/nuiiid arc Iriau'jU'K {Art.
508, 2) .
2, Tlie haufi of a pyramid is apnlijgoii (Art. 50S, 2).
634. A triangular pyramid is a pyramid whose base is
a triangle; a quad angular py am'd 'a '1 lose
base is a quadrila
A triangular pyram a d b U
faces. All these fa a ay miuya
taken aa tha basf.
635. A regu ar pyram d
is a pyramid who b
regular polygon, ud fo
of whose altitu
with the wiiLcr cf h b
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378 BOOK VII. SOIJD GEOJIirSRY
636. properties of a regular pyramid inferred immedi-
ately.
1. T]te hiieral ei^yc.'i of a rajidur pymmkl are equal, for
they are oblique lines drawn from a point to a plam;
cutting off eqnal distances from tlie foot of the per-
pendicular from the point to the plane (Art. 518) .
2. The lateral fwes of a regular pi/ranud are cqiiaj igifs-
celes Mangles.
637. The slant height of a regular pyramid is the ahl-
tude of any one of its lateral faci's.
The axis of a regnlar pyramid is-its aititode.
638. A truncated pyramid is thu portion of a pyramid
included between the base and a section cutting all the
lateral edges.
639. A frustum of a pyramid is the /fMT^
part of a pyramid included between the //M^-lw
base and a place parallel to the base. \gl.-...>J"'
The altitude of a frustum of a pyramid is
the perpendicular distance between the planes of its bases.
640. Properties of a frustum of a pyramid inferred Im-
mediately.
1. the lahral faces uf a frasiam of a pyramid are
ira peloids.
2. The lateral facet! of a frnglnm of a rerjulur punnnid
are equal isosceles trapesoids.
641. The slant height of the frustum of a regular pyra-
mid is the altitude of one of its lateral faces.
Ex. 1. Sliow that tlie foot of tlie nltitude of a regular pyramid
eoineides with the center of the cirele eircumscribed abont the base.
Ex. 2. The perimeter of the midaeetlon of the [rustum of a pyra-
mid equals oue-iuilf tha aura of the potimutm'a of tlie bases,
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rvRAMiDS 379
Pboposition XIV. Theorem
642. The lateral area of a regttlar pyramid is equal io
half the product of (he slant height hij the perimeter of the
Given 0~ABCI)F a regukr pyi'amkl with its lateral
area denoted by S, slant heiglit by L, and perimeter of its
base by P.
To prove g=J£XP.
Proof. The lateral faces OAJi, O/iO, etc., are equal
isosceles A. Art. GoG, 3.
Hence each lateral face has the same slant height, L.
:. the area of each lateral face = -} L X its base.
.-. the sum of all the lateral faees = i L X sum of bases.
^hLX P.
:. *S' = i LX P. A^,8,
643. CoK. 77te Intend urea of
thefrustnm of a regular puraviid is
equal io one-half the mm of the
perivieters of its iases multiplied
by its slant height.
Ex. Find tlie lttterii.1 ii
lieifj'lit is 32, uiid an l-S
uvea liXici,
I of whosi; base ia lli. Piad tho 1
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]iOOK VII. SO"LtI) OEOJIETltY
Pkopositiom XV. Theokem
644. If a pyramid is cut by a j)?D«e paruUi'l to the fi«se,
I, The lateral edges and the altitude are divided pro-
poriionalbj;
II. Tke section is a polygon similar to the base.
Given the pyramid S-ABCDF, with the altitude SO cut
by a plane MN, which is parallel to the base and intersects
the lateral edges in a, b, c, d, f and the altitude in o.
So
To prove I. — =^
11.
8 A SB SC
The section
ABCDF.
SO
abcdf similar to the base
Proof, I. Pass a plane through the vertes S II MN.
Then SA, SB, SC . . . SO, are lines intersected by three
I planes.
Sa _Sb _Sc _ ___ S
SA SB SO
SO
Art. 539.
II.
{Wiiy f )
ab II AB.
.: A Sab aud SAB are similar.
In like manner the A Sbc, Scd, etc., are similar to the
SBC, SCD, etc., respectively.
\sb)^
AB
~BC Ksc)
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oSrf/
ABCDF
SA^SO "
«6"
cihcif .
. &>'
PYRAMIDS 381
That is, the homologous sides of abcdf and ABCDF ave
proportional.
Also ^afjc = Z ABC, Ihal = I BCD, etc. Art. 538.
.•. section aferf/ is similar to tlio buse ABCDF. Art. 32!.
Q. E. D.
645. OoE. 1. A section of a pyramid parallel to the
hose is to the base as the square of its distance from the
vertex is to the square of the altitude of the pyramid.
But Ji:^^^ =
ABGDF SO-
646. CoE. 2. If two pyramids having equal altitudes
are cut hy a plane parallel to their bases at equal distances
from the vertices, the sections have the same ratio as the
bases.
Let S-ABCDF and V-PQE be two pyramids cut as
described.
n,, abcdf Ho' , , , par Yi
Then ^— =;=r- Art. G45 ; also -i-i— = ^^- WhjT
ABCDF «o- DQR VT
But rr=SO, and n = So. (Why?)
. »&«?/ ^ W ohrdf_ABCDF
•■ABCDF PQli °^ pqr PQR ^ " '
647. Cor. 3. If two 2>yrHmids have equal altitudes and
'equivalent bases, seetions iiiiide hy phnies parallel to the bases
at equal distances f rum Ihv verlk-vs art equivuhnt.
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KOLID (iEOMETRY
PhOI'OSITION XVI. TlIKOREM
648 Tlifi lohaiic of k tn mgitlai j^yrmmd is the limit of
i Hin "I the loliiwcs of n sciipf of inscribed, or of a series
f iKtni''' hIkcI ;;iisws of qtml altitude, if the number of
hi. tndif.miibj inueat,€i^
Given the triaogiilar prism 0-ABG with a series of in-
scriijed, and also a series of circnmseribed prisms, formed by
passing planes which divide the altitude iuto equal paris, aud
by making the sections so formed first upper bases, then
lower bases, of prisms limited by the next parallel plane.
To prove O-ABG the limit of the sum of each series, if
the number of prisms in each be indefinitely increased.
Proof, Each inseribad prism equals the circumscribed
prism immediately above it. Art. 629,
.■. {sum of circnmseribed prisms) — (sum of inscribed
prisms) = lowest civcumscribed prism, or ABC-K.
If the number of prisms be indefinitely increased, the
altitiide of each approaches zero as a limit.
Hence volume ABG-K=0, Art. 253, 3.
{for Us base, ABC, is cojistant Khile Us altitMe = ) .
.■, (snm of circumscribed prisms) — (sum of inscribeil
prisms) = 0.
.■. volume 0~ABC — (either series of prisms) ^ 0.
{for this difference < (ajyereiice between the two series, vihick
lust difference = 0).
,■, 0-ABC is the limit of the sum of the volumes ot
either series of prisms. q. e. b.
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Proposition XVII. Theorem
649. If two triangular pyramids have equal altitudes
and equivalent bases, tJiey arc equivalent.
Given the triangular pyramids 0-AEG and 0'~A'B'C'
having equivSreQt bases ABC and A'Ji'C, and equal
altitudes.
To prove O-ABC'o.O'-A'B'C.
Proof. Place the pyramids so that they have the com-
inou altitude R, and divide H into any convenient nnm-
ber of equal parts.
Through the points of division and parallel to the plane of
the bases of the pyramids, pass planes cutting the pyramids.
Using the sections so formed as upper bases, inscribe a
series of prisms in each pyramid, and denote the volumes
of the two series of prisms by V and I".
The sections formed by each plane, as KLM and K'l/M',
are equivalent. Art. 6*7.
.'. each prism in 0-ABC o corresponding prism in
O'-A'B'C (Art, C29). .-. F= F. As:. 2.
Let the number of parts into which the altitude is
divided be increased indefinitely.
Then V and V become variables with (h-ABC and
0'~A'B'C' aa their respective limits. Art. MM.
But I'oT'' always. (WLv?)
:, 0-ABC~0'-A'B'C'. (Why!)
(1. X. D.
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BOOK V)I. SOLID
PHOl'DSITrON XVIII. TlIEORF,:«
650. Thr vohinK' of a fridDditlar pyramid i
oiif-lhird Ihr iirodiid of iV.v bam: hy il^ nUilmh'.
Given the triangiilar pyramid 0-AB€, having its volnmo
denoted bj- V, the area of its base by B, and its altitude by H.
To prove V^^BXM.
Proof. Ou ABC as a base, with OB as a lateral edge,
construct the prism ABG-BOF.
Then this prism will be composed of the original pjTa-
mid 0-ABG and the quadrangular pyramid 0-ADFO.
Through the edges OJ) and OC pass a plane intersecting
the face ^DFC in the line B<7, and dividing the quadrangular
pyramid into the triangnlar pyramids 0~ADC and OSFC.
Then 0-AT)C^O-I>FC. Art. 649.
{for IkeijMrc the common rerlci 0,<i,nd the equal bases ABC andBFC).
But 0-DFC may be regarded as having as its vertex
and DOF as its base. Art. 634.
.-. (}'DFC=:^0-ABC. Art. 619.
.'. the prism is made up of three equivalent pyramids.
.-. (}-ABC=h the prism. As. 5.
But volume of prism = B X H. Art. 627,
.-. 0-ABC, or r=iBX H. Ax. 5.
g. E. B.
Ek. FJ^id the volume of a trmngula,r pyramid whosa altitude 13 12
ft,, ami whose b:iao is an equilateral triangle with a bide of 15 ft.
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Proposition XIX. Theorem
851. The volume of any pyramid is equal to one-third
iAe product of itn base by Us ultitude.
Given the pyraraid 0- ■il>( VF ln\ mg its volumfi denoted
^>y V, the area of its baae bj £., and its altitude by H.
To prove F= i B X E.
Proof. Through any lateral edge, as OD, and the diago-
nals of the base drawn from its foot, as AS and BD, pass
planes dividing the pyramid into triangular pyramids.
Then Y, the volume of the pyramid 0-ABCDF, will
Bqnal the sura of the volumes of the triangular pyramids.
But the volume of each A pyramid = J its base X E.
Art. 650.
Hence the sum of the volumes of A pyramids=i sum of
their bases X E. Ax. 2.
= k I) X E. Ax. 8.
.-. r=h BX S. Ax. 1.
Q. E. D.
852. OoR. 1. 5'fie volumes of two pyramids are to each
'>iher as the products of tkeir banes atid altitudes; pyramids
fiaving equivalent bases and equal altitudes are equivalent.
658. Cob. 2. Pyramids having equivalent bases are to
^eh other as their altitudes; pyramids having equal alti-
'Mifes are to each other as their bases.
854. Scholium, The volume of any polyhedron maybe
J'^und by dioidiuff the polyhedrmi Into pyramid.'^, finding the
^hnie of each pyramid •jcpundchj, and tailing their sum.
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BOOK VII. aOI.ID GEOMETRY
Proposition' XX, Theorem
656. The frustum of a triangular pyramid is eguivaleKt
to the sum of three pyramids whose common altitude is the
altitude of IM frustum, and tvJioae bases are the lower base
the upper base, and a mean proportional between the two
bases of the frustum.
Given ABG-DEF the frustum of a triangular pyramid,
having the area of its lower base denoted by B, the area
of its upper base by b, and its altitude by B.
To prove ABG-DEF ^ three pyramids whose bases are
B, h and V Bb, and whose common altitude is H.
Proof. Through E and AG, E and DC, pass planes divid-
ing the irustum into three triangular pyramids. Then
1. E-ABC has the base B and the altitude H.
2. E-DFG, that is, G-DEF, has the base b and the
altitude if.
3. It remains to show that E-ADC is equivalent to a
pyramid having an altitude H, and a base that is a mean
proportional between B and 6.
Denoting the three pyramids by I, II, III,
I_ A ABE __AB^_AC^ ^ ADC ^11
II A AI)t:~DE BE A I>FG III'
(Arts, 653, 391, 644, 321. Let the pupil supply the reaaon for each
I II
step in detail).
(Ax. l.)orII = l/lXiny
.-. E~ADG=y'{^HXB) (hSX b}=iHVBXb.
Hence, J.£OZ'iJ#— sum of three pyramids, as deseribecU
^. £■ S.
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656. Formula for volume of frustum of a triangular
pyramid. F=i JI {B+h + VmY.
Proposition XST. Theorem
657. The volume of Oie fi-usium of nny pyramid is
eguivaJent to the sum of the volumes of three pyramids, whose
common attitude 4s the altitude of the frifsium, and whose
lases are the lower base, the upper base, and a mean propor'
tional between the two bases of the frustum.
Given the frustum of a pj-ramid Ad, having its volume
denoted by V, the area of its lower liase by B, of Its upper
^>ase by 6, aud its altitude by H.
To prove T=h E (B + b + VBb). .
Proof. Produce the lateral faces of Ad to meet in K.
Also construct a triangular pyraiuid with base PQR
■qnivalent to ABCDF, and in the same plane with it, and
■Tith an altitude equal to the altitude of K-rABCDF. Pro-
luce the plan€ of. ad to cut. the second pyramid in pqr.
Then pqr'cahcdf. Art. 647^
.-. pyramid ST-ABCPFo pyramid T-PQIi. Art. (552.
Also pyramid K-abtdf = pyramid T-fqr. (Why ?)
Sttbtractiug, frustum Ad ^ frustum Tr. Ax. 3.
But volume Pr^-]. 11 Hi Jrh+VlThh
:. volume Ad^;, II {n ]-b^VBb).' (Why?)
-"Q. EiB. -
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SOLID GEOMETRY
Piioposrnox XXIL Tiieorkm
658. A truncuted triangular priim is vqvivalent to tTie
sitvi of three pumHiids, of which the base of the prism is the
common base, and whose vertices are the three vertices of the
inclined section.
Given the truncated triangiilai- prism ABG- PQR.
To prove ABC-PQR ^ the sum of the three pyramids
P-ABC, Q~ABC and Ti-ABC.
Proof. Pass planes through Q and AC,Q and PC, divid-
ing the given figure into the three pyramids Q-ABC,
Q-APG and Q-FBG.
1. Q-ABG has the required l*ase and the required
vertex Q.
2. Q~APC^B-APG, Art.652.
(/or they liave tkesame base, AFC, and the same altitude, their ver-
iices beind in a line |i bate APC),
But B~APG may be regarded as having P for its ver-
tes, and ABG fov its base, as desired. Art. 634.
3. Q-PBC<i= B-ARC (see Fig. 2). Art. 652.
\foT the base ASC ^ base PSC (Art. 300); and the altitudes of the iao
pyramids are eqnal, the vertices Q and B being in line || plane
FACB, in lahich the bases lie).
But B~ARG may be regarded as having R for its ver-
tex, and ABC for its base, as desired. Art. 63*
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-•. ABC-PQB o sura of three pyramids whose eommoii
base is ABG, and whose vertieea are P, Q, R.
659. Cor. 1. The vobinie of a truncated right triangii-
iar prism (Fig. -i) '.v '■q/'id to the product of its la^ie hij
lyne-third the auin of Ur lateral eOges.
660. Cor. 2. The vohime of any truncated triangular
prism (Fig. i) is equal to the product of the area of its
ngM section (ii/ one-third the sum of its lateral edges.
P8ISMAT0IDS
661, A prismatold is a polyhedron
bounded by two polygons in parallel planes,
called bases, and by lateral faces which are
either triangles, trapezoids or
662. A pfismoid is a prismatoid in
*bich the Itiises have the same number of
sides and liavt- their corresponding sides
parallel .
&"■ The Tolurae of a truncated right parallelopiptd equals the
•fea ol the lower base multiplied by oue-fourth the sum of tlie lateral
^i^eea (or by a perpend'ciilar from the ceuter of the upper base to tUe
lower basul.
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r,i:inn-ri'i{Y
Proposition XXIII. TiiKOiiEJi
The vohmie of a prisituifold is iq'int (o
the product of its ultiiude hy tin: si
fowr Hmes the area of its midsection.
sixth
of its bases and uf
Given the prisiuiitoitl ABCI'-FOK, with bases B and h.
midsection Jf, volume 1', atul altitude Tl.
To prove Y^\H {B-^h^A 21) .
Proof. Take any point in tlie midsection, and through
it and each edge of the prismatoid let planes be passed.
These planes will divide the figure into parts as follows :
1. A pyramid with vertex 0, base ABCD and altitude
i H, and whose volnme .". = ,V 77 X jR. Art. 65] ,
2. A pyramid with vertex 0, base FGK, and altitude i
if, and whose volume .■. =i 7/ X h. (Why?)
3. Tetrahedrons like O-A'lKl who^e volume may be
determined as follows (see Fig. 2):
AB^-l-FQ. (Why!)
.-.A .it?B=4 A PGQ. Art. 398.
.-. O-AGB^i 0-PGQ. Art. 653.
But 0~PGQ (or G-FQO) = h I'QO X h H = i HXPQO.
.: O-AGB^i SXiAPQO. (Why?)
.-. the sum of all tetrahedrons like 0-A(3B=i HXiM.
:. Y=-\ EXB + i HXh+i 11X4 M.
Or V=}E iB+h + iM). l!.S.l>.
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RFGULAE POLYHEDRONS
REGULAR POLYHEDRONS
664. Def. a regular polyhedron is a polyhedron all of
whose faces are equal regular polygons, and all of whose
polyhedral angles are equal. Thus, the uuhe is a regular
polyhedron .
Proposition XXIV. Theorem
665. Btit five regiihir polyh^dronti are p
Given regular polygons of 3, 4, 5, etc., s
To prove that regular polygons of the same number of
sides ean be joined to form polyhedral <^ of a regular
polyhedron in but five different ways, and that, conse-
quently, but five regular polyhedrons are possible.
Proof. The sum of the face A of any polyhedral angle
< 360°. Art. 583.
1. Each Z of an equilateral triangle is 60°. Art. i.ii,
3 X 60°, 4 X 60° and 5 X 60° are each less than 360°;
bat any larger multiple of GO° = or > 360°.
.". but three regular polyhedrons am be formed with
equilateral &. as faces,
2. Each Z of a si/Krtc* contains 90°. An. irii.
3 X 90° is less than 300°, but any larger multiple of
90" = or > 360".
.*. but one regular polyhedron enn be formed with
Squares as faces.
3. Each i^ of a regular jiewiai/oM is 108°. Art. \--i.
3 X 108" is less than 300°, but any larger multiple of
108° > 300°.
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Jii)^ HOOK VII.' SOLID riEo:Mi:'niv
.■. but one regular piil.\ lu'dmn u;ui lu' formed wHli regu.
lar peutagons as faeos,
4. Eaob Z of a regular iu'xaffoii is 120°, anil 3 X 120"
.". no regular polyhedron eau be formed with hexagons
or with polygoiiB with a greater number of sides as faces,
.'. but five regular polyhedrons are
666. T!ie construction of the regular polyhedrons, by ths
nse of cardboard, may be effected as folltjws;
Draw on a piece of cardboard the diagrams given below.
Out the cardboard half through at the dotted lines and eu-
tirely through at the full lines. Bring the free edgea
together and keep them in their respective positions by
some means, snch :is pasHnft- strips of paper over tViem.
TTfM
P^^ W'i^p -z^
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POLYHEDRONS ii))iS
POLYHEDRONS IK SEHESAL
PliOPOf^ITION XXY. TllEOKEM
667. III. inn/ polnhedron , the nitnibi'r of edges increased
by two eqiuils the number of vertices increused hy the miinher
offacen.
Given the polyhedron AT, with the number of its ver-
tices, edfjes and faces denoted by V, E and F, respectively.
Toprove E + 2=V+F.
Proof. Taking the shigle face ABCD, the nuraber of
edges equals the number of vertices, or E= V.
If another face, CRTT)., be annexed (Fig, 2), three new
edges, 6'i?, BT, TI), are added and two new vertices, B and T.
:. the mimber of edges gains oue on the aumber of ver-
tices, or E^ T+ 1.
If still another face, BQRC, beflunexed, two new edges,
BQand Qli, are added, and one new vertex, Q. .: E= V+2.
With each new face that is annexed, the number of
edges gains one on the number of vertices, till but one
face is lacking.
The last face increases neither the number of edges nor
of vertices.
Hence number of edges gains one ou number of vertices,
for every face except two, the first and the last, or gains
F~1 m all.
.■. tor thi- entire tiK'n-c, /■:= I'-f- F—'l.
That is £.' + 12= t'-i- t\ A.V. 3.
Q. s. n.
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.i'Ji ]!OOK vn . SOLID GEOJIETEY
Proposition XXVI. Theorem
668. The sum of Ike fncc amjlex of any j'ohjhedron
equals four right ant/Us iaknn us inunij times, less two, (o,-
the polyhedron has vtiikfs.
Given any polyhedron, with the sura of its face angles
denoted by S, and the number of its vertices, edges and
faces denoted by V, E, F, respectively.
To prove 8= { V—2) 4 rt. A .
Proof. Each edge of the polyhedron is the iatersection
of two faces, ,■. the number of sides of the faceE = 2 E.
:. the sum of the interior and exterior A of the faces =
2 S X 2 rt. A, or EX i rt. A . Avt, 73.
But the sura of the exterior A of each faee = 4 rt, A .
:. the sum of exterior A of the Ffaces = i'X4 rt. A.
Subtracting the sum of the exterior A
from the sum" of
all the A,
the
Bujn of the interior A •
of the P faces =
(.SX4rt.
A)-
-{FX4 rt. A).
Or
li={E-F) 4rt. A
But
E + 2=r+F.
Art. 6QT.
Hence
E~F^V~2.
As. 3.
Substituting
■,tor E—F, S=(V~~2) 4 r
t. A . Ax. 8.
Q. I. B.
E5, Veri
f J the laal two tlieorema iu the uase
of the cube.
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COMPARISON OF POLYHEDKONS
COMPARISON OF POLYHEDKONS. SIMILAR POLYEEDKOWS
Proposition XXVII. Theoreji
669. // two tetrahedrons have a trihedral angle of one
equal to a trihedral angle of the otJter, they are to each other
as the products of the edges including the equal trihedral
angles.
Given the tetrahedrons O-ABC aud O'-A'B'C, with
theirvohiines denoted by Faad V, respectively, and having
the trihedral A and 0' equal.
V OAXOBXOC
To prove yr ^yA' X O'B' X O'C'
Proof. Apply the tetrahedron O'-A'B'C to (h-ABC so
that the trihedral Z (y shall coincide with its equal, the
trihedral Z 0.
Di-aw CP and O'F 1 plane OAB, and draw OP the
projection of OC in the plane OAB.
Taking OAB and OA'B' as the bases, and OP and G'P'
as the altitudes of the pyramids 0-OAJi and C'-OA'B',
respectively.
V A OAB X CP A OAB ^ CP
But
A OA'B' XC'F A OA'B'
A OAB OA X OB
-Xr
07"
Art, <ir>'>
OA' X OB'
CP _ OC
A OA'B'
In the similar rt. A OCPand OC'F,
T^ OAX OBX OC ^ OA X OB X OC
■'■ V" OA' X OB' X OC O'A' X O'B' X O'C'
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?,nc,
SOLID GEOMETKV
670. Dep. Similar polyhedrons are polyhedmus having
the saaie iinmbei' of (aces, similar, each to each, and simi,
lai-ly placed, and having thcii- eorrespondiog polyhedral
augles equal.
PKOi'OSiTio^- XXVni. ThEORG-M
671. Any iiro s-iimlar polyliedrons way he (leeompoaed
into the t:ftme niimle.f of lef raited rons, similar, each to each
and similarly placed.
Given P and P', two similar poiyhfidrons.
To prove tlv;it P and P may be decomposed into the
same number of tetrahedrons, similar, each to each.
Proof. Take R and W any two homologous vertices of
P and P'. Draw homologous diagonals in all the faces of
P and P except those faces which meet at H and H', sepa-
rating the faces into corresponding similar triangles.
Through H and each face diagonal thus formed in P,
and through S' and each face diagonal in P', pass planes.
Each corresponding pair of tetrahedrons thus formed
may be proved similar.
Thus, in E-ABC and II'-A'B'C, tlie A EP,A and
H'B'A' are similar. Art. 329.
In like manner A HBC and H'B'C are similar; and &■
ABC and A'B'C are similar.
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SIMILAR POLYGONS 397
., NA /HB\ EC /BC\ AC . „.
.-. A ARC and A'WC .-u-e simiiar. Art, 326.
Hence the corresponding faces of H-ABC&mlH'-A'B'C
are similar.
Also their homologous trihedral li are equal. Art. 5S4.
.". tetrahedron S-ABC is similar to H'-A'B'C. Art. 670,
After removing S-ABG from P, and H'-A'B'C from
P', the remaining polyhedrons are similar, for their faces
are similar, and the remaining polyhedral A are equal.
As. 3.
By continuing this process, P and /" may be decom-
posed into the same number of tftrahedroos, similar, each
to each, and similarly placed,
Q. E. D.
672. Cor. 1. The hoDwloijoim vdyes of similar polyhe-
drons are proportional;
Any two homologous line'i in iiro aimilar polyhedrons
iiave the same ratio ws any other tiro homologous Hwh.
678. Cor. 2. Any tiro liiiimloijous faces of iiru .similar
polyliedrons are to each other «« the nquaret: of any tivo
homologous edges or lines;
The total areas of any tiro similar polyhedrons are to
each other as the sguares of any two homologous edges.
Es. 1. Iiithefigure.p.39,i,if tl
and those meeting at 0' ave *, 6, :
of the tetrahedrons.
Kz. 2, If the linear dimeasions of one room are twice as great as
the eorrespondiag dimensions of another room, how will their enrf aces
(and ,". cost of papering) compare 1 liow will thi'iv volumes compare T
Ex. 8, How many 2 iu. cubes can be cut from a 10 in, cube t
Ei. 4, If the hasL's of a prisraoid are raetansl^s whose dimensions
we 0, b a:id b. a. and altitude is E, And the formula for the volume.
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674.
each otii
BOOii YII. SOLID GEOMETRY
PROPOSITION XXIX, Thbokeji
Till', volume:; of tiro similar tdmhcdrons
r ns the cull's of any pair of Jioniologous
Given the similar tetrahedrons O-ABC and O'-A'S'G'.
To prove — — -.m:-,'
1" O'A''
^ . y OAX OBXOG ,^, „,
V (yA' X O'B' X O'C
^OA y OB y OG
O'A' O'B' O'c'
OA ^ OB __ PC
O'A' O'B' O'G'
V OA yO^y^ OA ^ ~0A^
' r'~0'A' O'A' O'A' ooT'^
El. 1. In the above figures, if AB=^2 A'B', fiud the ratio of, V
to r. Find the same, if AB^U A'B'.
Ex. 2 The meaBUremeDt of the volume of a, regular triangular
prism reduces to the laeasurBment of the lengths of how many straight
lineB ! of a fmstHin of a regular square pyramid ?
Ex. 3. Show how to eonstruet out of pasteboard a iBgular pli^m,
B puallelapiped, aad a truncated siiua.re prUtn.
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similar polygon's s
Proposition XXS. Theorem
675. The volumes of any two similar polyhedrons i
to each other as the cubes of .any two homologous edges,
of any other two homologous lines.
Given the polyhedrons AK and A'K' having their vol-
umes denoted by Fand V, and RB and B'B' any pair of
homologous edges.
To prove — "^ "•-.:".•:'
Proof. Let the polyhedrons be decomposed into tetra-
hedrons, similar, each to each, and similarly placed. Art. 671.
Denote the volumes of the tetrahedrons in Pby t-i, vs,
i-'i . . . and of those in J" by f'l, v'2, v'-i . . .
Then
V':
1 H'B'
Also
(/"•
each of
these rat
''°'~Wb'
j)
Art.
674, As. 1.
" t
n + t
'1 +
«'. + -
- ^\'
that is,
V
.^1
Art. 3la,
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iW HOOK Vil. SOLID GLO^TE'l'in:
TWKOitEMS rdXCEHNJNC POLVHEOROXS
Ei. 1. TliB Iriteral faees of :i rigbt prism are rectangles.
Ex. 2. A diagonal plaue of ti prism is parallei to every lateral edge
of the prism not coiitnined in the plnue.
Ex. 3. The diagouais of a parallelepiped bisoci^ each other.
Ei. 4. The square nf a diagonal o£ a reijtiingular parallelepiped'
oqaaJs the sum of the squares of the thrive edges meeting at a vertex
Ei. 5, Each lateral face ol :\ prism is pavallel to every lateral edgo
not contained in the face.
Ex.
lateral
6. Every section of a prism made by a plane parallel to a
Bilge is a parallelogram.
Ex.
allel ti
prism
7. It any two diagonal planes o'
D each other are perpendicular t
is a right prism.
the base of the prism, the
Ex.
Iflse is
8. What part of the volume of
. a face of the cube and whose vei
3 cube is the pyramid
■fex is the center of the
whose
onbef
Ex. fl. Any section of a regular squ
throngh the asis is an isosceles triangle.
are pyramid made by a
plane
Ex. 10. lu any regular tetrahedron,
the perpendicular from its foot to any !';
an altitude equals three
times
Ex.
1 1. Id any regular tetrahedron,
, ail altitude equals the i
.urn of
the perpendiculara to the faces from any point within the tetratedron.
Ex. 12. Find the simplest formula for the lateral area of a trun-
cated regular prism of n sides.
Ex. 13. The sum of the squares of the four diagonals of a paral-
lelepiped is equal to the sum of the squares of the twelve edges.
[Sno. Use Art. 352.]
Ex. 14. A parallelopiped iij symmetrical with respect to what point?
Ex. 15. A rectanglar parallelo piped is symmetrical with respect to
how many planes? (Let the pupil make a definition of a figure syni-
metrical with respect to a plane. See Arts. 4S6, 487.)
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EXERCISES OX POLYHEDHOXH -JOl
Ei. 16. The volumfl of a pyr'amid ivbose lateral odges mn the
three edges of the parallelepiped m«eting at a point is what part of
the volume ot the parallelopiped 1
Ex. 17. If a plane be passed through a vertex of a cube and tlie
tliagonal of a faae oot adjacent to the vertex, what part of the volume
of the enbe is eoataioed by the pjrainiil so formed !
Ex. 18. If the angles at tho vertex of a triangular pyramid are
right angles and each lateval edge equals a, show that the volume of
Es. 19. IIoiv large is a dihedral anglo at the base of a regular
pyramid, if tho apotheia ot the base equals the altitude of tho pyramid 1
the lateral surface.
Ex. 21. The section of a triaiiguiar pyrainiil by a plane parallel
to two opposite edges ia a parallelogram.
If the pyramid is regular, what kind of a parallelogram does tha
section become ?
Ex. 22. The altitude of a regul;
ot the base into seguients whii.'h ari
:ir tetrahedron
divide!
Ex. 23. If the edge of a reguh
elant height is ^ ; and hence th
ir tetrahedron
at the altitude
'* 3 '
ume is ^^'
Ex. 24. If the midpoints o£ all the edges of
B- tetrahedron except two opposite edges be
joined, a parallelogram is formed,
Ex. 25. Straight lines joining the midpoints
'^t the opposite edges of a tetrahedron meet in
s point and bisect each other.
Ex. 26, The midpoints ot tl^- ed-es of a reg'
^e vei'.ic..^ ot a regulur octiihedron.
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JIOOK Vll, SOLID GEOMKJ'HV
EXERCISES. CROUP Ca
intOELKMS CONrKRSING POLYHF.DBONfi
isect the volume ot a given [insm by a plane parallel U
Ex. 3. Through a given point pass a plane which shall bisoct the
volume of a given paralleloplpeil.
Ex. 4. Given an edge, couatruet a regular tetrahedi'ou.
Ex. 5. Given an edge, eonstruct a regular octahedron.
Ex. 6. Pass a plane tiirougb the axis of a regular tetrahedron eo
that the section shiili bo an iaoaceles triangle.
Ex. 7. Pass -J, plane through a cube so that the section nhall be a
regular hexagon.
Ex. 8. Through three given lines no two o£ which are parallel
pass planes which shall form a parallelepiped,
Ex. 9. Prom cardboard eonatiuct a regular square pyramid each
o£ wbose faces is an equilateral triangle.
EXERCISES. GROUP 69
REVIEW F.XEKCISES
Make a 1
ist of the pr
operties of
Ex. 1.
Straight li
nos in space.
Ex. 9, Eight prisms.
Ex.2.
One line a
nd one plane
Ex. 10. Parallelopipeds in gen-
Ex.3.
Two or n
lore lines an
,d ^'■'''■
one
plane.
Ex, 11. Rectangular parallelo-
Ex.4.
Two plane
B and one lin
e, pipeds.
Ex.5,
Two plane
sand two line
g Es, 12, Pyramids in general.
Ex.6,
Polyhedrons in general.
Ex. 13, Regular pyramids.
Ex, 7.
Similarpolyhedrons.
Ex. 14 Frusta of pyramids.
Ex, 8.
Prisms in
general.
Ex, 15, Truncated prisms.
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Bdox VIIT
CYLINDERS AND CONES
CYLiHDEKS
876. A cylindrical surface ii5
a carved surface generated bj a
straight line which moves so :in
constantly to touch a given fixed
curve and constantly be pLirLilL-l
to a given fixed stniigiit line
Thus, every stiadow cast by a pomt
of light at a great distance, ao l>y a
Btar or the sun, appTOximatPs the
oylindrioal form, that is, is, Ijouaded Cjimdrnal surface
by a, oylindrieal Hurfaoe of light. Henee, in all radiations (ns of light,
heat, magDetism, etc.) from a point at a great distance, we are
concerned with oylindrieal snrfacea and solids.
677. The generatrix of a cylindrical surface is the mov-
ing straight line; tlie directrix is the given curve, as CDE;
an element of the cvlindrical suffice i'; the moving straight
line m any one of its positions, as BV
678. A cylinder lo i solid bounded bj
^ cylindrical sniface md L\ two piialiel
planes.
The bases of a c\lmdei ul its puallel
plane faces; the lateral surface is the
cylindrical surface iiiclud d liLt\\eeu the
parallel planes foiming its bT^f s tlic alti
tude of a eyliudHi is the distiin,e lictwteu the 1
The elements of a e^lmhi ait the elements of the
cylindrical surface boundm': it.
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iO[
JiOOK Yin, SOLID GEfniKTKY
679. Property of a cylinder inferred immediately. All
the elements of a cylinder are equal, lor they ;ire parallel lines
included between parallel plai.u's (Arts. 533, 076),
Tlie cjiinilers most important in practical lite irt, those determined
by their stability, the ease with wliich tlipy can bi, iiude from com-
680. A right cylinder is & cyhndci
wliose elements are perpendicular to tlie
bases.
681. An oblique cylinder isoue-nliui
elements are oblique to the bases,
682. A circular cylinder is a cjlmd i
whose bases are circles.
683. A cylinder of revolution is a c\liii-
der generated by the revolutiou of a lect
angle about one of its sides as an axis.
Hence, a cylinder of revolutioa is a right
circular cylinder.
Some of the properties of tills solid ore derived
most readily Tiy cousidering it as generated by a re-
volving rectangle ; and otliers, by regnrdiag it an a
particular kind of cylinder derived from t!ie grnLial
684. Similar cylinders of revolution ne f\lmdeib gen-
erateil by similar ructaiigles revolving ibout homologous
sides,
685. A tangent plane to a cyliud \ is t, plane whioh
contains one element of the cylinder, and whi-'h do^!^ not cut
tlie cylinder on being produced.
Es. 1. A plane passing through a taugent to the base of a circu-
lar cylinder and the element drawn through the point of contact is
tangent to the cylinder. (For it it is not, etc.)
Ex. 2. If a plane is tangent to a dicular cylinder, its intersection
villi tlic lilane of the base is tangent to the base.
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CYLINDERS
686 A prism inscribed lu a cylinder is a pr
lateral edge'^ ue ebmeiits of the cylinder, and wnos
ni I H m tlur- 1 I es of the cylinder.
405
prism whose
687 A prism circumscribed about a cylinder is a prism
whose litenl fice-. are tangent to the cylinder, and whoso
bases are polygons eii-enmscribed about the bases of the
cylinder.
688. A section of a cylinder is the figure formed by the
intersection of tlie cylinder by a plane.
A right section of a cylinder is a section formed by a
plane perpendicular to the elements of the cylinder.
689. Properties of circular cylinders. By Art. 441 the
area of a circle is the limit of the area of an inscribed or
circumscribed polygon, and the circumference is the limit
of the perimeters of these polygons; hence
1. The volume of a circular cylinder is the limit of the
volume of an inscribed or circuiAscribed prism.
2. The lateral area of a circular cylinder is the Umil of
the lateral area of an inscribed or circumscribed priism.
Also, 8. By metJiodf loo advanaedfor this book, iC may be pravfd Ih-il
the jierimeUr ol'u rig/it atttuin is the limit of the perimetm- of a rif/hl ner-
tvm of an insrvib&l or drcinnMri'ied pri»'n.
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690. Ei-enj section of a rylUuli-r mcuh hi) a plane pass-
ing tkrou'jh an ekmcid is « parnllrloarain.
Given the <>ylindpf AQ cut by a ■plixuc passing through
the element AB and fonriiiig the section ABQP.
To prove A BQP a CJ .
Prooi. API! BQ. Art. 631.
It reraaiES to prove that PQ is a straight line II AB.
Through P draw a line in the cutting plane ll AB.
This line will also lie in the cylindrical surface. Art. 676.
.■. this line must eoineide with FQ,
(fnr the fine drawn lies in both the cutting phmc and the cylindrical
surface, liencc, it iiiiiKt lie their iatvrscction).
:. FQ is a straight line || AB.
:. ABQP is a /ZJ . (Why?)
(J. E. D.
691. Cor. Every section of a rirjht cylinder made, by a
plane parsing through an element is a rectangle.
Ex. 1. A doov siTiii!;iii;; on its hinges generates what kind of a
Es, 2. Every sectioQ of a paBiillelopipeii maJe by a plane iater-
secting all its la,teral edges is a imrullciogiaw,
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CYLINDEBS
Proposition If. Theorem
The bases of a cylinder are equal.
Given the cylinder AQ with tlie bases AVB and CQB.
To prove base APB = base GQD.
Proof. Let AC and BD be anj' two fixed elements in
the snrface of the cylinder AQ.
Take P, any point except A and B in the perimeter of
the base, and through it draw the element PQ.
Draw AB, AP, PB, CD, CQ, QI).
Then ACand Bi> are = and II. (Why!)
.-. AD is a ^17 . (Why?)
Similarly AQ and BQ are 07 .
.-. AB^CD, AP^CQ, and P.P=T)Q. (Why?)
.■. AAPB=A CQD. (Why?)
Apply the base APB to the base CQD so tl.at AM coin-
cides with CD. Then P will coincide with ^,
(for A AFB = A CQD).
But P is any point in the perimeter of the base APB.
.: every point in the perimeter of the lower base will
coincide with a corresponding point of the perimeter of
the upper base.
.'. the bases wiil coincide and arc equal- Art. 47.
^, E, B.
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408 HOOK Vm. fiOT.lD fJT:0:METl!Y
693. Cob, 1. The sscfioit.'i of a cjlinder madp by two
paraJlfl phities cutting all the elements are equal.
For the sections thus formed ai'e the bases of the eylinder-
iucluded between the cutting planes,
694. Cor. 2. Attn section of a cyHmhr paniUel to the
base is equal lo the bat^e.
pROPOsiTiON III. Theorem
695. TJie lateral urea of n cirmdnr vylimler is equal fo
tlie product of the perimeter of a ritjltt section of the cylin-
der by an elemertf.
Givea the circular cylinder AJ, haviu" its lateral area
denoted by 6', an fiknifnt by JS, and the perimeter of a
right section by P.
To prove S=PXE.
Proof. Let a prism with a regular polygon for its base
be inscribed in the cylinder.
Denote the lateral area of the inscribed prism by 8',
and the perimeter of its rijjht section by P'.
Then the lateral edge of the inscribed prism is an ele-
ment of the cylinder. Constr
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4ft. asa, 3.
(Whyf)
(Why?)
Q. E. D.
CYLINDKKS 409
.-. S'^PXE. Art. COS.
If the number of lateral faces of the inscribed prism be
indefinitely increased,
iS" will appi'oaeh ^ as a limit. Art. 689, 2.
P' will approach P as a limit. Art. 689, 3.
And P'Xfiwill approach PXE as a limit.
But. ,S" = PX.E always.
.-. 8 = PXE.
696. Cor. 1. The laieml area of a cyliwhr of involu-
tion is equal I" the produci of Ihe civionferpnee of its base,
by its altHiirh.
697. Formulas for lateral area and total area of a cylin-
der of revolution. Denoting the lah-ral area of a cylinder of
revolution by S, Ike total area by T. the radius by R, and
the altitude by U.
T=2 TtRH + -2 nR" :. T^2 nR {H^ R).
Ex. 2. If the a
of the Lase (H = l
terms of li t also, in tenns nf // f
Ex. S. What do they liecorae, if the altitude eqwals tlie diametsr
<rt the base ?
Ex. 4. Ill a cyliniier of revolution, what is the ratio of the latersl
"^ea to the area, of the base f to the total area !
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4H) BOOK VTII. SOLID OLOMKTP.V
I'HOropmuN i\. TiUiORicM
698. The. vnlums of a riycular cylinder is equal to the
product of its base hij Hh tdlitudc.
Given the eircnlav I'yli'iLler AJ, having its volume
denoted by V", its base by B, and its altitude by K.
To prove r=BXM.
Proof. Let a prism having a regular polygon for its
base be inscribed iu the cylinder, and denote the volume of
the inaeribed prism by T', and its base by B'.
The prism will have the same altitude, H, as the
cylinder.
.-. V'^B'XH. (Why?)
If the number of lateral faces of the inscribed prism
be indeflaitely increased,
V will approach V as a limit. Art. 689, i,
B' will approach B as a limit. (Whj!)
And B'X H will approach BXH asd. Hmit (Wh?))
But V'^B' X ff always. (Why?)
.-. V=BX B. (Whys)
Q. E. a.
699. Formula for the volume of a circular cylinder.
By use of Art. 450,
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CYLINDERS 411
Proposition V. Theorem
700. The lateral areas, or the total areas, of hco simi-
lar cylinderi' of revolution are to each other as the squares
of their radii, or as the squares of their altitudes; and
their volumes are to each other as the cubes of their radii,
or as the cubes of their idfitudcs.
Given two similar cyliiidura of revolution having their
lateral areas denoted by S and S', their total areas by T
and T', their volaraes by T^and P, their radii by R and
B', and their altitudes by Sand H', respectively.
To prove S -. H'^T: T^BT- : B'^-^E" -. B'^\
and y : y' = E? : R'^ = E'' : S'\
_ H+R
U'~R' R' + R''
2 rcitB BXE _R ^,H ^R" ^g"
Proof.
Arts, Z'il, 309,
S'
r
I TiR'H' R' X B'
T,y~rF-
ir-
(Wliy?)
I 7tR' {W + R') B' ff' + E' B'-
(Whyt)
(Why f)
Q. E. B.
Ex. If a eylindrlcal c
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CONES
701. A conical siirface is a sur-
face giaiierated by ii straight iiin'
which moves so as constantly to toucli
a given fixed curve, and constantly
pass through a given fixed point.
Tb-tts every shado\r ciist by a ne p
of iigM iaeonifliil ic foi'iu, tlmt is, is
by a conical sui'faee of liglit. Hen li
study ot conical surfaees and solid s a
portant from the fant that it eone n a
oases of forces radiating from a neai p n
702. The generatrix of a
surface is the moving straight
AA'; the directrix is the give f d "
carve, as ABC; the vertex i tixed |ont as 0; an
element is the generatrix in anv o e ot t po tons as BB'.
703. Tiie upper and lower nappes oC a conical snrface
are the portions above and belon' the vertex, respeetivelv
g,s 0-ABCi\w\ O- A' B'C.
Usually it is coavenicat lo limit a conical surface to a single nappe,
704. A cone is a solid bounded by a
conicai snrface and a plane cuttmg all the
elements .
705. The base of a eone is the face
formed by the cutting plane; the lateral
surface is the bounding coincal surface;
the vertex of the cone is the vertex of the i
conical surface; the elements of the cone
are the elements of the eonieal surface; cone
tlie altitude of a cone is the perpendicolar distance from .the
vertex to the plane of tho base.
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706. A circular cone is i\ con?, whose base is a circle.
The axis of a circalm- cone is the line drawn from the
vertex to the center of the base.
707. A right circular cone is a c I one 1 e i
is perpendicular to the plane of the 1
An oblique circular cone is a cLri, 1
whose axis is oblique to the base.
708. A cone of revolution is a co e ge e
rated by the revolution of a rifjlit t
about one of its legs as an axis.
Hence a cone of revolution ai 1
circular cone are the same solid. ■>
709. Properties of a cone of revolution inferred im-
mediately.
1. The altitude of a con^^ of rrmhiiion U the <uis of the
cone.
2. All the elements of a vone ,.f rri'<>]i>/i^n are equal.
710. The slant height of a cone of revolution is any one
of its elements.
711. Similar cones of revolution are uones gciienited by
similar right triangles revolving about homologous sides.
712. A plane tangent to a cone i^ a pliine which con-
tains one element of thu uouu, but wliich does not cut the
conical surface on being produced.
Ek, 1. A plane passing througii a taageut lo llic ba.se of a circular
one and the element drawu through the poiat of coutiict is tnngt-nt to
Es, 2. ]f a plane is li
the plane of Iho base i-, t;
s intersection with
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713. A pyramid inscribed m
a cone is a pyramid whose liitci i!
edges are elements of the coi e
and whose base is a polygou in
scribed in the base of the none
714. A pyramid c i r c u m
scribed about a cone is a pjia [_ _,
mid whose lateral faces are tin
gent to the cone and whose base is a polygon eireumseribed
about the base of the cone.
715. Properties of circular cones. By Art, 441 the area
of a circle is the limit of the area of an inscribed, or of a
circumscribed polygon, and the circumference is the limit
of the perimeters of these polygons; hence
1. The vohmie of a cir<;nlar cone is the limit of the vol-
ume of an inscribed or eireumseribed pyramid.
2. The lateral area of a circular cone is the limit of the
lateral area of an inscribed or circumscribed pyramid.
716. A frustum of a cone is the \ oi
tion of tlie cone iuelndcd between th bi e
of the eoiic and a plane parallel tl
base.
The lower base of the frustum
base of the cone, and the upper
the frustum is the section made
plane parallel to the base of the cone.
What must be the altitude and the lateral surface of a
frustum of a cone; also the slant height of the frustum ol
a cone of revolntion ?
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CONES 415
Proposition VL Theorem
717. Every section of ti cone made hy a plane passing
through its vertex is a triangle.
Given tlie oone S-AFBQ with a plane passing through
the Tertes S, and making tlie section SPQ.
To prove SPQ a triangle.
Proof. PQ, the intoi-section of tlie base and the cutting
plane, is a straight line. (Whj!)
Draw the straight lines SP and SQ.
Then 8P and HQ must be in the cutting plane; Art. 498,
And be elements of the (lonical surface. Art. 701.
,*. the straight lines SP and SQ are the intersections of
the eonieal surface and the euttiug pliuie.
.', the section SPQ is a triangle, Art, 81.
{/or )I is bomuled bij Ihrc: nhviglit lines).
Ea. What liiiid of triaufjie
Blade bj a plaxie tiivough the >.
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BOOK VIII. KOLiD IIEOMETKY
Proposition VII. Theorj^m
718. El-cry section of u cirnilar voite made by a plane
^parallel to the buse is a civl''.
Given the circular eoue SAB with nph a section made
by a plane parallel to the base.
To prove apb a circle.
Proof. Denote the center of the base by 0, and draw
the axis, SO, piercing the plane of the section in o. '
Through SO and any element, SP, of the conical stir-
face, pass a plane cutting the plane of the base ia the
radius OP, and the plane of the section in op.
In like manner, pass a plane through SO ami SB form-
ing the intersections OB and oh.
:. OrWop, and OB]\ob. (Why?)
.-. A SFO and 8B0 are similar to A Spo and Sbo,
respectively. Art. 328,
■' OP
But OP^OB. (Why J)
.'. op = ob. {Why t)
.". oph is a circle. (Wlyt)
Q. E. ».
719. Cor. The axis of a circular cone passes through
the center of every section parallel to the base.
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CONES 417
Proposition VIII. Theorem
720. The lateral arm of a cone of revolution is equal to
half the product of the slant height by the drcumferenee of
the base.
Given a cone of revoltilion having its taterat ares de-
noted by S, its slant lieight by L, and the cireiiinforeuce
of ita base by C.
To prove S^iCXL.
Proof, Let a regular pyramid bo eireuinscribed about
the cone.
Denote the lateral area of the pyramid by S', aad the
perimeter of its base by P. .
Then S'^iPXL. Art. 612.
H the number of lateral faces of the cireumserlbod
pyramid be indefinitely increased,
S' will approach S as a limit. Art. 7i5, 2.
r wilt approach C as a limit. Art. 44i.
And J P X L will approach J CXL as a limit. Art. 253, 2.
But S'^i PX L always. (Why?)
.-. S^^ CXL. (Whyf)
Q. £. n.
721. Formulas for lateral area and total area of a cone
of revolution. Denoting the radius oi' the base by R,
.S'=i {2 Ttfl X L) :. ti=7TRL.
Also T=7iiiL +71^- :. T^TcU {L + B).
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BOOK V!II. SOLID (iEOMETRY
PkOPOSITIOX IX. TUEOREll
722. The volume of a circular cone is equal to one-thin
of the product of its base by its altitude.
Given a circular cone having its volume denoted by V
its base by B, and its alfcttude by E.
To prove V^iBXH.
Proof- Let a pyramid with a regular polygon (or its
base be inscribed in the given cone.
Denote the volume of the inscribed pyramid by F, and
its base by B'.
Hence V = ^B'XH. Art. gsi.
If the number of lateral faces of the inscribed pyramid
be indefinitely increased,
V will approach F as a limit, (Why ?)
B' will approach B as a limit. (Why ?j
And i B' X H will approach J B X S" as a limit. ( Wby i)
But V'^i B' X ff always. (Why?)
r^iBXH. (Why?)
q. £. i>.
723. Formula for the volume of a circular c
Ex. 1. If, in a cone of revolution, if =3 and JJ = 4, find S,r and F,
Ex. 2. II the altitude of a cone of revolution eijuals the radius of
the base, what do the formulas for S, T and V become 1
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CONES 419
Proposition X. Theorem
724. The lateral areas, or the total areas, of two simi-
lar rones of revolution are to each other as the squares of
their radii, or as the squares of their altitudes, or as the
squares of their slant heights; and th^ir volumes are to each
ether as the cubes of these lines.
Given two similar cones of revolution having their
lateral areas denoted by S and 8', their total areas by T
and P, their volumes by Y and V, their radii by jB and
E', their altitudes by H and H', and their slant heights by
L and I/, respectively.
To prove S -. .^'=^T -. T' = IP: R'- = W' -. H'" = J? -. L'"-;
and V : V' = E? : E''' = H' : H'^ = L^ : L'K
K V L' + E''
Proof.
(Vfhy?)
(WhjT)
725.
such thf
triangle.
(Why?)
Q. E. D.
Dek. An equilateral cone is a eone of nivolution
^. a section through the axis ia au equilateral
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i2i) BOOS VIII. SOLin GEOJlETRi'
Proposition XL Theoeem
726. The lateral area of a frustum of a cone of re.voli
tioH is equal to one-half the sum of the circumferences of i
bases multiplied iy Us slant height.
Given a frustum of a cone of revolution having its
lateral area denoted by S, its slant height by L, the rarlii
of its bases by B and r, and the circumfereuces of its bases
by C and c.
Toprove S=i (C+c)XL.
Proof. Let the frustum of a regular pyramid be cireum-
scribed about the given frustum. Denote the lateral area
of the eireumscribed frustum by S', the perimeter of the
lower base by P, and the perimeter of the upper base by p.
The slant height of the circumscribed frustum is L.
Hence S'-J {P + p) X L. Art. 643.
Let the pupil complete the proof.
(J. E. B.
727. Formula for the lateral area of a frustum of a cone
of revolution. S~i {2 TtiJ + 2 tit) L.
.-. S^n {U + r)L.
728. Cor. The lateral area of a frustum of a cone of
revolution is equal to the product of the circumference of Us
Midsection by its slant height.
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CONES 421
Proposition XII. Theorem
729. The volume of the frustum of a circular cone is
equivalent to the volume of three cones, whose common alti-
tude is the altitmle of the frustum, and whose bases are the
lower hase, the upper base, and a mean proportional between
the two iases.
Givea a fnistura of a eircalar cone having its volume
denoted by V, its altitude by H, the area of its lower base
by B, and tiiat of its upper base by b.
To prove V = h S {11+ b + Vb X h).
Proof. Let the frustum of a pyramid with regular poly-
gons for its bases be inscribed in the given fmstum.
Denote the volume of the inscribed frustum by V, and the
areas of its bases by B' and 6'.
.-. V' = h S iB' + 6' + l/ff^xT') , (Why t)
If the number of lateral faces of the inscribed frustum
be indefinitely increased, V will approach V, B' and V
approach B and b respectively, and B' X V approach B X
b, as limits. Art. 719.
Hence, also, B' + b'+V^B' X b' will approach B -\~ b +
V'b X J as a limit. Art. 353.
But V'= iH {B + J'+V-^' X b') always. (Whj?)
.■. V=iE{B + b + l/fiXb). (Why!)
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i'12 ];00K VIII. SOIJI) GEOMETRY
730. Formula for the volume of the frustmu of a circii,
lar cone.
F= h if (TtA'- + Tir' + VtiH-'x ni^) .
Ex. 2. If a eocieal oil-can is 12 iu. iiielii liow much more tin is
required to make it tliaa to make a airailai' oil-pan in. high ! How
much more oil wiil it hoid ?
Ei. 3, The linear dimeneions of i eonieal funnel are three timt-s
tbose of a similar funnel. How mxieh more tin is required to walie
the first f How much more liquid will it hold f
El. 4. Make a similar oompariaon of iiylindrieai oil-tanks. Of
EXERCISES. CROUP 70
THBORBMS COKCERNISQ CYLINDERS AND CONES
El. 1. Any section ot a cylinder of revolution through its axis is
a rectangle.
El. 2. On a cylindrical surface only one straight line can be
drawn through a given point,
[Sua. For if two straight lines could bo drawn, etc.]
Es. 3. The intersection of two planes tangent to a cone is a
straight line through the rertex.
Ex. 4. If two planes are tangent to a cylinder, their line of inter-
seetion is parallel to an element ot the eyliuder,
[SuG. Pass a plane -L to the elements of the cylinder.]
Ei. 5. If tangent planes be passed through two diametrically
opposite elements of a circular cone, these planes intersect iu a
straight line through the Tertes and parallel to the plane of the base,
Ex. 6. In a cylinder of revolution the diameter of whose base
equals the altitu<3e, the voiumc equals one-third the product of the
total surface hy the radius of the base.
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EXERCISES ON THE CYLINDER AND CONE 423
Ek. 7. Aeylinder and a cone o! revolutiou have the same base
Bnd the same altitude. Ficd tbe ratio of their lateral surfaoes, and also
of their voluroes.
Ex. 8. It an equilateral triangle whose Bide U a be revolved iiljout
ooe of ita Bides as an asis, find (he area generated in terms of a.
Ex. 9. If a rectangle whose aidea are a and b be revolved first
about the aide a as an axis, and then about the side b, find the ratio of
the lateral areas generated, and also of the volumes.
Ex. 10. The baaoe of a cylinder
ooacentric. The two solids have th
ot the base of the cone is twice the <
def. What kind of Hue is the iute
and how far is it from the base 1
it a cone of revolution are
attitude, and the diameter
er of the base of the cjlin-
a of their lateral surfaoea,
Ejc. 11. Determiue the same when the radius of tha «■
imea the radius of the cylinder. Also when r times,
Ex. 12. Obtain a formula in terms o£ r for the volume of the
'rustum of an equilateral cone, in which the radius of the upper base
a r and that of the lower base is 3r.
Ex. 13. A regular hoxagon whose aide
oual through the center as axis. Find, i
and volume generated.
folvea about a diag-
Ex. 14. Find the locus of a point at a giv
en distance f
straight line.
Ex. 15. Find the locus of a point whos
e distance f
line is in a given ratio
to ita distance froc
a a fixed pis
dioular to the line.
Ex. 16. Find the lo
us of all straight 1
nea which m
angle with a given line
at a given point.
Ex. 17. Find the lo
cus of all straight 1
nea which m
angle with a given plan
e at a given point.
Ex. 1 8. Find the locus of all points at c
tanee from the surface of a given cylinder of
n the surface of a gvv
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JK vni. SOLID GEOJIIM'KY
EXERCISES. QKOUP 71
)N'Cf.KN'IXG Till'; l'VLlSr>F.[l AND CONE
Ex. !. Through a given element o( a tireular eyliader, paaa v
plane l.augent to tbe ejlicder.
Ex. 2. Through a given element of a circular ciine, pass a plane tniij^Li
Ex. 3. About a given circular oylind
s regular polygon for it^ base.
Ei. i. Through a giren point outside a circuUr cylimier, paas a
plane tangent to the cyiinder.
Ex. B. Through a given point outside a given elnmlar eone, pas3
a plane tangent to tbe cone.
[St'ti. Through tbe vertei of tlio cone and the given point pass a
line, and produce it to meet the plane of the base.]
Ex. 6. Into what eegiuents must the altitude of a coneof revoliitirjn
he divided by a plane parallel to the base, in order that the volume of
the cone be bisected?
Ex. 7. Divide the lateral surface of a given cone of revolution into
two equivalent parts by a plane parallel to the base.
Ex. 8. If the lateral surface of a cylinder ot revolution be ont
along one element and unrolled, what Movt of a plane figure is formed f
lience, out of cardboard eonatruot a cylinder of revolution with
given altitude and given eireumterenee.
Ex. 9. If the
literal ^
uitice
of ac
one of
revolution be
out along
one element and
umoHed
what
sort ol
: a. plai
16 figure is formed !
Hence, out ol
taidboa
li toi
JStiuet
a com
e of revohition
ot given
slant height
Ex. 10. C(n=
trmtan
e^nlla
teril CI
TOO ont
of pasteboard.
Ex. 11, eout
itruit a 1
luatliu
. Of a<
■one of
revolution out
, of paste-
board.
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Book IS
THE SPHERE
731. A sphere is a solid bounded by a surface all
points of which are equally distant from a point withiu
called the center.
732. A sphere may also be defined as a solid generated
by the revolution of a semicircle about its diameter as an
axis.
Some oE the properties of a epheio may be obtained more readily
from one o( the two deflcitions given, and some from the otliec.
A sphere is named by naming tiie point at its eecter, or by naming
three or more points on its surfaae.
733. A raduis of a sphere is a line drawn from the
center to any point on the surface,
A diameter of a sphere is a line drawn through the
center and terminated at each end by the surface of the
sphere.
734. A line tangent to a sphere is a line having bnt one
point in common with the surface of the sphere, however
Jar the line be produced.
(425)
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4::f) BOOK IX. SOLID GEOJffiTRY
735. A plane tangent to a sphere is ;i plane having bu
one point in common with tlifi surface of the sphere, how-
ever far the plane be produced.
736. Two spheres tangent to each other are spheiv--
whose surfaces have one poiut, and only one, in common.
737. Properties of a sphere inferred immediately.
1. All radii of a sphere, or of equal spheres, are equal.
2. All diameters of a sphere, or of equal spheres, are.
equal.
3. Two spheres are equal if their radii or their diame-
ters are equal.
PROi'OSiTioN I. Theorem
738. A section of a sphere, tnade by a plane is a circle.
Given HI 1 1 1.) (i I id PCD a section made by a plane
cutting the sphere.
To prove that FGD is a circle.
Proof. From the center 0, draw OA X the plane of
the section.
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THE SPHERE 427
Let Che B. fixed point on the perimeter of the section,
and P any other point on this perimeter.
Drawee, AP, OC, OP.
Then the A OAP and OAG hvb rt. A. Art. 505.
OP=OG. (Whj!)
OA^OA. (Why!)
A OAP=A OAC. (Whjf)
.-. AP=AC. (Wiiy t)
But P is any poiut on the perimeter of the section PCD.
:. every point on this perimeter is at the distance AO
from A.
.". PCD is a circle with center A. Art. 197.
q. E. s.
739. Cor. 1. Circles which are se.ctions of a sphfre
wade by planes equidistant from the center are equal; and
conversely.
740. CoE. 2. Of two circles on a sphere, the one nuide
by a plane more remote from the center is smaller; and
conversely.
741. Dep. a great circle of a sphere is a circle whose
plane passes through the center of the sphere.
742. Dep. A small circle of a sphere is a circle whose
piane does not pass through the center of the sphere.
743. Dep. The axis of a circle of a sphere is the
diameter of a sphere which is perpendicular to the plane of
the circle. Thus, on figure p, 42fi, BB' is the axis of PCD.
744. Dep. The poles of a circle of a sphere are the
"istremities of the axis of the circle. Thus, P and B', of
■igure p, 426, are poles of the circle PCD.
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428 BOOK i\. SOLID f;EOMETKY
745. Properties of circles of a sphere inferred imiiif
diately.
1. 2'Ae axiK of a circle of a sphere pnxses through tl
center of the circle; and conversely.
2. Parallel circles have the same axis and the same poln
3. All great circles of a sphere are equal.
4. Every great circle on a sphere bisects the sphere ai,
its surface.
5. Two great circles on a sphere bisect each other.
For the line of intersection of the two planes of th
circles passes througli the center, and hence is a diametf'
of each circle.
6. Through two points (not tite extremities of a diameio
on the surface of a sphere, one, and only one, great circle ch
he passed.
For the plane of the great circle nmat also pass throng
the center of the sphere (Art. 741), and through thro
points not in a straight line only one plane can be passe
(Art. 500) .
7. Through any three points on the surface of a sphen
not )rt the same plane icith the center, one small circle, a/t
only one, can he passed.
746. Def. The distance between two points on the sui
face of a sphere is the length of the minor arc of a grt-a
circle joining the points.
Ex. 1. If the raiii.is of a sphere
is 13 in-, find the vadlua c
circle on the sphere made by & plane
at a distance of 1 ft. from
center.
Ex.2. Wliat geographical eirclea
00 the earth's surface are gn
and what small circles t
Ex. 3. What is the largest nuiabt
ir of points in which two cir
ou tlie surface of a aphevo can iutersi
Eictf Why?
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THE SPIIEKE 41!y
Proi'OSition II. Theorem
747. All points in the circumference of a circle of a
sphere are eqvnlhi lU^tnut from each pole of the circle.
Given ABC a circle of a sphere, and P atid P' its poles.
To prove the arcs FA, PB, PC equal, and ares F'A,
r'B, P'G equal.
Proof. Draw the chords PA, PB, PC.
The chords PA, PB and PC are equal. Art. 518.
,■. area FA, PB and PC are equal. Art. 313.
la like manner, the arcs F'A, F B and FC may be
Q. E. B.
proved equal.
748. Def. The polar distance of a small circle on a
sphere is the distance of any point on the circumference of
the circle from the nearer pole.
The polar distance of a great circle on a sphere is the dis-
tance of any point on the eircumfereuce of the great circle
from either pole.
749. Coil. The polar distance of a great circle is the
quadrant of a great circle.
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430 BOOK IS. SOLID GE0\£TE¥
Proposition III. Theoreji
750. // a point on the surface of a sphere is at a quaii
rfinf's dhhmce from two other poinl.i on the. surface, it i.
the pole of the great circle through those points.
Given PB and PC quadrants on the surface of the
spliere 0, and ABC a, great circle through B and C.
To prove that P is the pole of ABC.
Proof. From the center draw the radii OB, OC, OP.
The arcs PB and PC are quadrants. (Why t)
.■- ^ POB and POO are rt, A. (WhyT)
.-. PO ± plane ABC. (Why?)
.". P is the pole of the great circle ABC. (Why i)
Q. 1. D,
751. Cor, Through two given points on the surface of
a sphere to describe a great circle.
Let A and B be the given points.
From A and B as centers, with a
quadrant as radius, describe arcs
on the surface of the sphere inter-
secting at P. With P as a center
»nd a quadrant as a radius, describe a great circle.
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THE SPHEliE
Proposition IV. Theorem
752. A plane perpendimlar to a radius at its extre
is tangent to the sphere.
Given the sphere 0, and the place MN J. the radius OA
of the sphere at its extremity A.
To prove MN tangent to the sphere.
Proof. Take P any point in plane MN except A. Draw
OP. Then OP>OA. (Whj-f)
.■. the point P is outside the surface of the sphere.
But P is any point in the plane MN except A.
.: plane MN is tangent to the sphere at the point A,
(for every point in Vie plane, except A, U milstde the surface fif t!ie sphere).
Art. 73&. a, B. D.
753. Cor. 1. A pkinc, or a line, which is tangent to a
Sphere, is perpendicular to the radius drawn to the point of
<:ontact. Also, if a plane is tangent to a sphere, a perpendicii-
far to the plane at its point of contact passes through the center
of the sphere.
754. Cor. 2. A straight line perpendicular to a radius
''J a sphere at its extremity is tangent to the i^phere.
755. Cob. 3, A straight line tangent to a circle of a
Sphere, lies in the plane tangent to the sphere at the point of
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4:62 BOOK IX. SOLID GKo:\iT:T]iy
756. Cob. 4. A straight line drawn in a tangent plan
and through the point of contact is tangent to the sphere :
that point,
757. Cor. 5. Two straight lines tangent to a sphere ,
a given point determine the tangent plane at that point.
758. Def. a sphere circumscribed about a poiyhedro
is a sphere in whose surface lie all the vertices of tli
iiolyhedron.
759. Df.p. a sphere inscribed in a polyhedron ij;
sphere to which all the faces of tlic polyhedron are tangent
PROPOSITION V. Problem
760. To circumscribe a sphere about a give
hedron.
Given the tetrahedron ABCD.
To circumscribe a sphere about ABCD.
Construction and Proof. Construet E and F the centers
of circles circumscribed about the A ABC and BGD, re-
spectively. Art. SSfi.
Draw ES X plane ABC and FK X plane BCD. Act. 5U.
Draw FG and FO to G the midpoint of BG.
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THE SPHERE Add
Then EG and FG are X BC. Art. ii3.
.-. plane EOF 1 BC. Art. 509.
.-. pljine EOF ± plane ABC. Art, 555.
.-. SH lies in the plane FOE. Art. 558.
In like manner FK lies in the plane F6E.
The lines EG and FG are not ll,
[for Ihfy nieel in the point G),
:. the lines Bff and -F£: are not !1. Art. 122.
Hence ES innst meet FK in some point 0.
But EJ/^ is the locus of all points equidistant from A,
B and C\ and F^ is the loeus of all points equidistant
from B, C and Z>. Art, 520.
,'. 0, which is in both EH and FK, is equidistant from
A, B, Oand D. (Whyl)
Hence a spherical surface constructed with as a center
and OA as a radius will pass through A, B, G and D, anil
form the sphere required, q, e. f.
761. Cor. 1. Four points not m Ike same plane deter-
mine a sphere.
762. Cor. 2. The four perpendiculars erected at the
'"'enters of the faces of a tetrahedron meet hi a point.
763. Cor. 3. The sije planes perpendicular to the edges
of a tetrahedron at their midpoints intersect in a point.
764. Dep, An angle formed by two curves is the angle
formed by a tangent to eaoh curve at the point of inter-
section.
765. Def. a spherical angle is an angle formed \iy
two interseetiutr avns of great circles on a sphere, and
l>euee by tangents to these arcs at the point of intersection.
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34 BOOK IX. POLID r.EOMKTRY
PkOPOSITION VI. PltOHLEM
766. To inscribe a sphere in a given ieffihrilrtm.
Given the tetraheilion ABOD
To inscribe a spheie m ABC1>
Coaatruction and Proof. BibPct the dihoilral angle D-
AB-C hy the plane OAB; similarly bisect tlie diliedral A
■whose edges are BG and AG hy the planes OBC and OAG,
respectively.
Denote the point in whieh the three bisecting planes
intersect by 0.
Every point in the plane OAB is equidistant from the
faces DAB and GAB. Art. 562.
Similarly, every point in OBC is ec|uidistant from the
two faces intersecting in BG, and every point in OAG is
equidistant from the two faces intersecting in AG.
.: is equidistant from all four faces of the tetra-
hedron. Ax. 1.
Hence, from as a center, with the J. from to any
one face as a radius, describe a sphere.
This sphere will te tangent to the four faces of the
tetrahedron and .'. inscribed in the tetrahedron. Art. 759.
767. OoR. The planes bisecting the s
of a tetrahedron meet in one point.
: dihedral angles
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THE SPHERE
Proposition VII. Problem
768. To find the radius of a given material sphere.
Given the material sphere 0.
To construct the radius of the sphere.
Constnictioa. With any point P (Fig, 1) of the surface
of the sphere as a pole, describe any" convenient circum-
ference on the surface.
On this eireumferenee take any three points A , B and C.
Construct the A ABC (Fig. 2) having as sides the three
chords AB, BC, AC, obtained from Fig. 1, by use of the
compasses. Art. 283.
Circumscribe a circle about the A ABC. Art. 286.
Let KB be the radius of this circle.
Construct (Fig. 3) the right A kpb, having for hypot-
enuse the chord pb {Fig. 1) and the base ii. Art. 384.
Draw bp' ± hp and meeting pk produced at p'.
Bisect /jp' at 0.
Then op is the radius of the given sphere.
Proof- Let the pupil supply the proof.
Q. E. F.
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436 BOOK IX. SOLID GEOTilKTRY
Proposition VI]!. Theorem
769. The intersection of itvo spherical surfaces is the
circumference of a circle whose plane is perpmdicularHo the
line joining the centers of the spheres, and whose center is i)i
that line.
Given two intersecting © and 0' wtiinh, by rotation
about tlie line 00' as an axis, generate two intersecting
splierical siirfaces.
To prove that tlie intersection of the spherical surfaces
is a ©, whose plane ± 00', and whose center lies in 00'.
Proof, Let the two circles intersect in the points P and
Q, and draw fhe common chord PQ.
Then, as the two given ® rotate about 00' as an axis,
the point P will generate the line of intersection of the two
spherical surfaces that are formed.
But PR is constantly 1 OO". Art. 241,
.', PB generates a plane J. 00' Art, 510,
Also PR remains constant in length.
,', P describes a circumference in that plane. Art. 197.
Hence the intersection of two spherical surfaces is a O ,
whose plane X the line of centers, and whoso center is in
the line of centers.
Q. E. B.
Tlie abovft demonstration is an illustration of the use of the second
definition of a sphere (Art. 732).
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THE SPHEKE 437
Peoposition IX. Theorem
770. A spherical angle is measured by the arc of a great
circle described from the vertex of the angle as a pole, and
included between its sides, produced, if necessary .
Given /.BAG a spherical angle formed by the intersec-
tion of the arcs of the great circles BA and €A, and BO
an are of a great circle whose pole is A.
To prove /. BAG measured by are BG.
Proof, Draw AI> tangent to AB, and AF tangent to
AC. Also draw the radii OB and OC.
Then Al> X AG. Art. 230.
Also OB L AO iJorABisaqm-iranl).
:. OB WAD, (Why?)
Similarly OGWAF.
:. ^B0(J= IDAW
But Z BOG is measured by arc BG.
:. IDAF, that is, LBAG, is measured by t
Art. 538.
Art. 257.
■■C BG-
(Why?)
Q. E. 9.
771. Cor. A spherical angle is equal in the plain
ofthv dihtdral angle formed hij (he planer of Us sides
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4r,S BOOK IX. HOLIU liEOJIETUY
SPHERICAL TRIANGLES AND POLYGONS
772. A spherical polygon is a portion of the
a sphere boumlyd by three or more
arcs of gi'eat circles, as ABGD.
The sides of the spherical polygon
are the bounding arcs; the vertices
are the points in which the sides in-
tersect; the angles are the spherical
angles formed by the sides.
773, A spherical triangle is a. splieriea! poiygon of three
sides.
Spherical triangles are classified in the same way as
plane triangles; viz., as isosceles, equilateral, scalene,
right, obtuse and acute,
774. Relation of spherical polygons to polyhedral angles.
If radii be drawn from the center of a sphere to the ver-
tices of a spherical polygon on its surface (as OA, OB,
etc., in the above figure), a polyhedral angle is formed at
0, which has an important relation to the spherical poly-
gon ABGD
Bach face angle of the polyhedral angle equals {in yium-
her of degrees contained) the corresponding side of the spheri-
cal polygon;
Dach dihedral angle of the polyhedral angle equals the
corresponding angle of ike spherical polygon.
Hence, corresponding to each property of a polyhedral
angle, there exists a property of a spherical polygon, and
conversely .
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THE SPHERE 4J9
Ilenoe, also, a trihedral angle and its parts correspond
to a spkfricnl triangle and its parts.
Of the common proportieB of a polyheilral angle and a epherical
polygon, some are discovered more readily from the one figure and
some from the other. In general, the spherical polygon ia simpler
to deal with than a polyhedral angle. For instance, if a triheUral
angle were drawn with the plane angles of its dihedral angles, nine
lines would be used, forming a complicated tigure in solid apace;
whereas, the same magnitudes are represented in a apherieal triangle
by three lines in an approximately plane figure.
On the other hand, the spherieal polygon, beeauao of its lack of
detailed parts, is often not so suggestive of properties as the poly-
hedral angle.
Proposition X. Theorem
775. The sum of tu'o sides of a spherical triangle is
greater than the third side.
Givea the spherical triangle ABC, of whieh uo side is
larger tliaii AB.
To prove AC-^ BO AB.
Proof. From the center of tlie sphere, 0, draw the radii
OA, OB, OC.
Then, in the trihedral angle O-AHG,
/LAOC. + IBOV > AOB. Art. 582.
.-. AV-^ BO > AH. Art. 774.
Q. E. 5.
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440 BOOX IX. SOLID CKOUETllY
776. CoK. 1. Any side of a spherical iriautjleis gfealn-
thin the. difference betiveen the other two mles.
777. Cor. 2. TJie shortest path ietween two points on
the surface of a sphere is the arc of the great circle joining
those points.
For any other path between" the two points may be
made the limit of a series of arcs of great oircies connect-
ing successive points on the path, and the sum of this
series of arcs of great circles connecting the two points is
greater than the single arc of a great circle connecting
them.
Proposition XI, Theorem
778. The smn of the siiits of a spherical polygon is less
than 360°.
Given the spherical polygon A BCD.
To prove the sum of the sides of ABCD < 360°.
Proof. From 0, the center of the sphere, draw the radii
OA, OB, OC, 01).
Then lAOB-\- IBOG+ lOOI>-\- IDOA < SCO".
(Why T)
.-. AB -^ B(] + CD -^ DA <360°. Art, 77*.
Q. E. I>.
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SPHERICAL TKIANGLES
441
779. Def. The polar
triangle of a given triangle
is the triangle formed by
taking the vertices of the
given triangle as poles, and
deserihing arcs of great cir-
cles. (Hence, if each pole
be regarded as a center, the radius used iu describing each
arc is a quadrant.) Thus A'B'C'is the polar ti'iangle of
ABC; also B'E'F' is the poiar triangle of BEF.
Proposition XII. Theorem
780. If one spherical triangle is the polar of another,
then the second triangle is the polar of the first.
Given A'B'C the polar triangle of ABC.
To prove ABC the polar triangle of A'B'C .
Proof. B is the pole of the arc A'C .
:. arc A'B is a quadrant.
Also C is the pole of the arc A'B'.
.'. arc A'C is a quadrant.
.*. A' is at a quadrant's distance from both B and C.
.*. A' is the pole of the arc BC. Art. 750.
Iq like manner it may be shown that B' is the pole of
AC, and C the pole of AB. g. E, 9.
Art. 779.
(Why t)
(Why?)
(Why?)
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442 I'.OOK !X. SOLID aEOlIETUY
Phoposition XIII. Theorem
781. Ill ii sjihfrical triangle <iiid its polar, each angle of
one Irianglc is the siipphmeiii of the side opposite in the
other li-iangle.
Art. tso.
(Whyf)
(Why!)
Given the polar ^ ABC ami A'B'C with the sides of
^£(7 denoted by a, b, c, and the sides of A'B'C denoted
by «', h', c', respectively.
To prove A + «' = 180°, B + 6'=180°, C' + c'=180°,
A' + a= 180°, B' + b= 180°, C + c - 180°.
Proof. Produce the sides AB and AG till they meet
B'C in the points D and F, respectively.
Then B' is the pole of AF :. arc B'F=90'
Also C is the pole of AD :. arc CD -90'
Adding, B'F + C'I>=180°.
Or B'F + FC + .BF^ 180°. Ax. 6-
Or 7J'C' + 7H'-180°.
But B'C'==a', and /ii'' is the measnre of the A A. Art, 770.
.-. A + a' -180°.
In like manner the other supplemental relations may be
proved as specified.
Q. a. B.
782. Def. Supplemental triangles are two spherical
triangles each of which is the polar triangle of the other.
This new name for two polar triangles is due to the
property proved in Art. 781.
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SPHERICAL TRIANGLES 443
Proposition XIV, Theorem
783. The sum of the angles of a spherical triangle is
greater than 180°, and less than 540°.
Given tliG spherical triangle ABC.
To prove J + S + C > 180" and < S40°.
Proof. Draw A'B'C, tlie polar triangle of ABO, and
denote its sides by a', S', c'.
Tiien A + a' - 180" 1
B -\- I' = 180" > Art. m.
C + c' = 180" '
.-. A -r B + C + a' + b' + c' = 540°. . . (1) Ax, 2.
Hut ^a' + b' + c' < 3ilO' Ai-i. -7R.
j „' 4- i' + c' > 0=
Subtracting eacb of these In tnni from (1),
A + B + C > 180° and < 540". ak. ii.
(I. E. B.
784. COH. A spherical triangle may have one, two or
three right angles; or it may have one, two or three obtuse
angles.
785. Def. a birectangular spherical triangle is a splier-
wal triangle containing two right angles,
786. Dei\ a trirectangular spherical triangle is a
spherical triangle containing three right angles.
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444
HOOK IX. SOLID GEOMETIiV
787. GoR. The surface of it sphere may he dimded into
fight inrectaiigiilar spherical triangles. For let three planes
X to each other be passed through the center of a sphere, etc.
788. Def. The spherical excess of a sphei-lciil t.ria
is the excess. of the sum of its angles over 180°.
igle
789.
whieh Y
Dep. Symmetrical spherical triangles are triangles
ive their parts eqnai, but arranged in reverse order.
Three planes passing through the isenter of a sphere
form a pair of symmetrical spherical triangles on opposite
sides of the sphere (see Art. 580), as & ABC and A'Ji'C
of Fig. 1.
790. Equivalence of symmetrical spherical triangles.
Two plane triangles which have their parts equal, but ar-
ranged inreverse p'
order, may be a^ /\
made to coincide / \v /^ \
by lifting up one / x. ^ ^ 'X X ^-'
triangle, turning
it over in space, and plaeiug it upon the other triangle.
But two symmetrical spherical triangles cannot be made
to coincide in this way, because of the curvature of a
spherical surface. Hetiee the equivalence of two sym-
metrical spherical triangles must be demonstrated in some
indirect way.
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SPHERICAL TEIANGLES 443
791. Property of symmetrical spherical triangles. Two
isosceles symmetrical spherical triangles are equal, for they
can be made to coincide.
Proposition XV. Theorem
792. Two symmetrical spherical triangles are equivalent.
Given the i
formed by plane
(See Art. 789.)
To prove
Proof. Let
the points A, B,
Also draw
great ®.
Abo P'
lymmetrical spherical A ABC and .I'R'C,
s passing through 0, the center of a sphere.
A AnCOA A'B-C.
P be the polo of a small circle passing through
Draw the diameter POP'.
PA, PB, PC, P'A', P'B', P'C, all arcs of
PA = PB t= PC.
--PA, P'B' ^ PB, P'C = PC.
Ax. 1
Similarly
And
Adding
In case the
A'B'C, V-X the
, ,'. P'A' = P'B' =P'C'.
J and P'A'B' are symmetrical isosceles
A PAG^^FA'C.
A PBC=^ PB'C,
l^PAB-+APAC+APBC
^AP'A'B'+/^FA'C'-\-AP'B'C'. kx.
AABC^AA'B'C. Ax.
polos /'and 7" fall outside the ^ ABC iiii
nuni! simply the demoiistratiou. q. e. b.
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BOOK IX. SOLID r.EOMK'niY
i'Hiii'o^rnoN XVI. Tii
793. On the same sphe?
{ingles are equal,
I. If two sides and the included angle of one are equal to
liro .sides atid the included angle of the oilier; or
JI. Ifiivo angles and the inehtded side of one a
io two angles and the included side of the other,
the corresponding I'quil parts being arranged in the saw/;
order in each case.
r on equal spheres, ivn ;»■;.
ire equitl
I. Given the spheriRai A ABC and J>EF, in which
AC=i>F, CR = FE, and ZC=IF.
To prove A ABC= A DEF.
Proof. Let the pupil supply the proof (see Book J,
Prop. YD-
II. Given the spherical & ABC and DEF, in whieh
Z(";=ZF, lJi = /.E, unA CB^FE.
To prove A ABG^ A J)EF.
Proof. Let the pupil supply the proof (see Book I,
Ex. 1.
tre 12 in.
If the
and 3
Ex. 2.
15 inciiea
The t£
In dmm
lineof centers of two spl:
ill,, liow ai'e tho spheres
auk on a motor car is a crylltider l!6 inches long and
leter. How many gallons of gasolene will it hold 1
equilateral cone, find the ratio of the lateral area to
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SPHEEICAL TRIASGLES
Peoposition XVII. Theorem
794. On the same sphere, or on equal spheres, two tri-
angles are symmetrical and equivalent,
I. If two sides and the inchtded angle of one are equal to
two sides and tlie included angle of the other; or
II. If two angles and the inrAuded side of one are equal
to two angles and the included side of the other,
the cotfesponding equal parts being arranged in reverse
order.
I. Given the spherical & ABC and DSF, in which AB
= DE, AC = DF, iind £A = £D, the eorrespomling parts
being arranged in reverse order.
To prove A ABC symmetrical with A DEF.
Proof. Construct the /MVE'F' symmetrical with ADEF.
Then A ABC may be made to coincide with A D'E'F',
Art, 703.
{haring two sides and the hiclmh-d L etiuat ami arraiigeil in llie
>ameor,hy).
But A D'E'F' is symmetrical with the A DEE.
:. A ABC, which coincides with A D'E'F, is symmet-
rical with A hEF.
ir. The second yiirt of the theorem
same way.
; proved in t!
I), E. D.
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BOOK IX. SOLID GEOMKTKV
Proposition XYIII. Theorem
795. If two triangles oti Ike same sphere, or equal
spheres, are mnhtally equilateral, they are also imitnally
equiangular, and therefore equal or symmetrical.
Given two mutually equilateral spherical A ABC and
A'B'C oil the same or on equal spheres.
To prove A ABC and A'B'C equal or symmetrical.
Proof. Prom and 0', the centers of the spheres to
■which the given triangles belong, draw the radii OA,
OC, O'A', O'B', O'C.
Then the face A at — corresponding face A at (y.
Art. 774.
Hence dihedral A at = corresponding dihedral A at 0',
Art. 584
,'. A of spherical A .4 £(7= homologous A of spherical
/\ A'B'C. Art. 774,
.". the A ABC and A'B'C are equal or symmetrical
ficcording as their homologous parts are arranged in the
same or in reverse order. Art. 789.
796. Note. Theconditions in Props. XVlaad SVIII which make
two sphei'iea! triangles equal aro the same as those which malse
two plane triangles equal Hence many other propOBitions occur in
Bpherieal geometry which are identical with correspondlag proposi-
tions in plane geometry. Thus, many of the construction problems of
epherioal geometry are Bolvd in. the sama way as the corresponding
construction problems m plane jfeomelryi as, to bisect a given
aogle, etc.
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'SPHEBICAL TEIANGLES 449
Proposition XIX. Theorem
797. If two Iriangles on the same sphere are mutually
equiangular, they ore idno mutually equilateral, and there-
fore equal or symmetrical.
Given the muhially equiangular spherical ii Q and Q'
on the same sphere or on equal spheres.
iitiiaOy equilateral, and
To prove that Q and Q" are i
therefore equal or symmetrioal.
Proof. Oc.nHtruct P and P' the polar & of Q and Q',
respective b .
Then A P aud P' are mutually equilateral. Art. 7S1.
.". iSv P and P' are mutually equiangular. Art. 795.
But Q is the polar A of P, and Q' of £". Art. 780.
.". A Q and Q' are mutually equilateral. Art. 78i.
Henee Q and Q' are equal or symmetrical, according as
their homologous' parts are arranged iu the same or in
:.'everse order. Art. 789.
Q. E. B.
798. CO!E. If tu-o mutualhj equiangular tiianijtes are
0)1 unequal uplfren, their correspomUng sides have the same
ratio as the radii of their respective spheres.
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450 liOOK [X. SOLID CEOMETliV
I'HOFOSITION XX. TlIEOlJEM
799. Jii an isosceles spherical triangle the angles opp,,
site the eqnul sides are equal.
Given the spherit-a! A ABC in whii^li An=-AC.
To prove Z7J=ZC.
Proof, Draw an arc from the vertex .1 to i>, the mid-
point of the base.
Let the pupil suppb' the roraaiiider of the proof.
I'KorosiTiON XXI. Theorem (Conv. of Prop. XX)
800. Jftivo angles of a spherical triani/le are equal, the
sides opposite these angles are equal, and the triangle is
isosceles.
Given the spherical A ABG in which ZB= Z 0.
To prove AB^AC.
Proof. Construct A A'B'G' the polar A of ABC.
Then A'C' = A'B'. Art
.-. ^6"= Z7J'. Art
.-. AB = AG. Art
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SI'IIEEICAL TRIANGLES 451
Proposition XXII. Theorem
801. 2m any spkt^rical triangle, if two angles are tm-
equaJ, the sides opposite these angles are unequal, and the
greater side is opposite the greater angle, and Conversely.
Given the spherical A ABC in which ,^. JiAO is greater
To prove liC > BA.
Proof. Draw the iirc AD making IPAO pqw.\\ lo 10.
Theu JiA = DC. Art,. 800.
To each of these equals add the are HI).
:. BD+ J)A^TiT>-\- T>(\ or liC. (Why?)
But ill A BBA, BI) + T)A > BA. (Wl.yT)
.-. BO > HA. Ax. 8,
Let ihe popii ijrnve the eonver.se by the indirect method
see Art. 106).
Q, £, D.
Ex. 1. Bisect a givi^n spherical aHgle.
Es. 2. Bisect B given are of a great circle oii a sphei'f-
(■'lii^il to a given spherical angla on tlie same sphere.
Ex. 4. Fimi (lie ioeiia ot the ccoterB o( tbe oirolea o£ a epher
foi'iiifd by pianos perpendloulac to a given ilinmeter of tlip give
'pheru.
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i'y2
ROLTD GEOMElliY
SPHERICAL AREAS
802. Units of spherical surface. j\
may be measured in tonus of, eirlicr
1. The customary units of arm, as
square foot, etc., or
2. Spherical degrees, or spherids.
803. A spherical degree, or spherid, is one-ninetieth
part of one of the eight trireetangnlar triaugles into which
the surface of a sphei-o may be divided (Art. 787), or -^is
part of the surface of the entire sphe
A solid degree i
spherical surface
a square inch, a
ie-nicet!eth part of a tri rectangular angle (se
lU).
I sphere
804. A lune is a portion of the surface of i
bounded by two semicii-eumferences
of great circles, as PBPC of Fig. 1.
The angle of a lune is the angle
formed by the semieircumfevences
which bound it, as the angle BPG.
805. A zone is the portion of the
surface of the sphere bounded by
two parallel planes.
A zone may also be defined as the sur-
face generated by an arc of a revolving
semicircumfetenpe. Thus, if QFQ' {FIb- 2)
generates a sphere by rotating about QQ",
its diameter, any arc of QFQ', as ISF,
generates a zone.
806. A zone of one base is a zone
one of whose boiitiding planes is
tangent to the sphere, as the zone
generated by the arc QtJ of Pig. 2. pig. -.-.
807. The altitude of a zone is the perpendiuulat- tiis
tance between the bounding planes of the zone.
The bases of a zone are the circumferences of the circles
of the sphere formed by the bounding planes of the zone.
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Spherical areas 4yd
Proposition XXIII. Theorem
808. The area generated hy a straiglit line revolving
about CHi axis in its plane is equal to the projection of the
line upon the axis, mtiUipUed by tM circumference of a circle
whose radius is the perpendicular erected at the inidpoint of
the line and terminated by the axis.
Given AB and X¥ in tlie same plane, CD the projection
of AB on XY. PQ the ± bisector of AB; and a surface
generated by the revolution of AB about XY, denoted as
"area ATi."
To prove area AB^ CD X 2 nPQ.
Proof. 1. In genera!, the surface generated by AB is
tlie surface of a frustum of a cone (Fig. 1).
,-, area AB=AB X 2 tiFR. am. 72S.
Dvaw AFLBD, then
A ABF and PQE are similar. Art. 32s.
.-. AB: AF^PQ: PR. (Why!)
.-. AB X P7^ = AZ'■'X PQ, or 071 X PQ. (Why!)
Substituting, area AB= CBX1 nPQ. Ak. 8.
2. If ABWXY {Pig. 2), the surface generated by AB
is the lateral surface of a cylinder.
.-. area AB- Ci> X 2 tiPQ. Art, ddl.
■S. If the point A lies in the axis XY (Fig. 3), k't the
pupil show that the same result iw obtained. q. e, d,
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4j4 book IX. SOLID GEOJIETRY
PiiO POSITION" XXR'. Theorem
809. The area of the surface of a sphere is equal io fht
prodiin of the diameter of the sphere by the circum fere.net
of a gnat virde.
Given a sphere generated by the revolution of the semi-
circle ACE about the diaiueter AK, with the surface of the
sphere denoted by S, and its radins by li.
To prove S==AEX2 nli.
Proof. Inscribe in the given semicircle the half of a
regular polygon of an even nnmber of sides, as ABODE.
Draw the apothem to each side of the semipoiygon, and
denote it by a.
From the vertices B, G, D draw Ji to AE.
Then area AK = AF X2na.\
areaflC=F0X2 -^a.
^ r, ,^j' ■., n 1 Art. 808,
area CB^0KX2 na.
&reaDE==KEX2 na. ]
Adding, area ABGBE^AE X 2 na.
If, now, the number of sides of the polygon be indefi-
nitely increased,
area ABODE approaches S as a limit. Art. m.
And a approaches E as a limit. (Why f)
.-. AE X 2 Tta approaches AE X 2 ttK as a limit. (Why !)
But area ABGDE^AE X 2 jt« always,
.-. S = AEX2nR. (Why!)
Q. S. B.
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SPHEEICA.L AREAS 435
810. Formulas for area of surface of a sphere^
Substitutmg for AE its equal 2 E, S-4 7ti^^
Also denoting tlie diameter of the sphere by jD, E = i J).
:. S=4 7l('~V, orS-TiD^.
<!)'
811. Cor. 1. The surface o/g sphere is equivalent to four
limes the area of a great circle of the sphere.
812. Gob. 2. The areas of the surfaces of two spheres
are to each other as the squares of their radii, or of their
diameters.
For, if ,S and S' denote the surfaces, E and B' the radii,
and D and ly the diameters of two spheres,
8 ^ 4 ^E^ ^ir- 8 ■htC- ^jy^
S' inR''^ B'2' ^'^^6" 7ii>'= D'-'
813. Property of the sphere. The following property of
the sphere is useil m tlm pi oof of Art. 809: //, in tlie generat-
ing arc of ami zoni. a biokcn Uiic be inscribed, whose vertices
divide the arc ndo equal parts, then, as tlie number of these
parts is increased indefinitely, the area generated by the broken
line approaches the area of the zone (is a limit. Hence
Cor. 3. The area of a zone is equal to the circumference
of a great circle multiplied by the altitude of the zone.
Thus the area generated by the arc BC = FO X 2 «.B.
814. Con. -t. On the same sphere, or on equal spheres, the
areas of Iwo zmn-ti are to each other as the altitudes of the
zones.
Ex. 1. Find Ihc iii-c;i nf a sphere wliosi? difimftcr !p 10 in,
Ex. 2. Fiiid tlif nni\ of a mue of alUtudir ;i io., on a sphere -whose
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liOOK IX. SOLID
PKorosiTioM XXV. Theorem
815. The area of a lime in to the urfa of (he svr~
face of ilie sjjhere as the angle of (he hine is to four riijht
Given a spliere having its area denoted by B, and on the
epiiere the inne ABCJ) ot £A with its area denoted by L.
To prove L: S=A° -. 360°.
Proof. Draw FB3, the great G whose pole is A, inter-
secting the bounding area of the lime in B and D.
Case I. When tlte arc BD and the circumference FB3
are commensurable.
Find a common measure of BD and FBH, and let it be
contained in the arc BD m times, and in the circumference
FBHn times.
Then arc BD : eirenmference FBII~m -. n.
Through the diameter AC, and the points of division of
the circumference FBH pass planes of great © .
The arcs of these great ® will divide the surface of the
sphere in » small equal luues, m of them being contained
in the lune ABCD.
:. L:8=m:n.
:. L : jS=arc BD : circumference FBH. (Why ?)
Or i ; «=.4° : 3G0°. Art. 257.
Case II, When the arc BD and the circumference FBH
are incommemurahle .
Let the pupil supply the proof. q. e. ».
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SPHERICAL AREAS 4 57
816. Fonnulafor the area of a lune in spherical degiees,
or sphends. The sucfaee of a sphere contains 720 Hpheriils
(Art. 803). Hence, by Art. 815,
L L spherids
"720 spherids
that is, IJie area of a hine in 32>]ier!cal degrees is equal (0
twice the number of angular degrees in the angle of the liute.
817. Formula for area of a lune io square units of area.
■4 7it{""-^m'
r„ or ;.=
818. Cor. 1. Oh ihe same Kphere, or on equal .''ijheres,
two htties (ire to mch other as their anijlex.
819. Cob. 2. Two bmes with equal angles, but, on mi-
equal spheres, are to each other as the squares of the radii of
their spheres.
'^F-A TtV'^A
For L : i'.~^ : -^. or /, : L' = B- , E-.
Ex. 1. Find the area in spherical degrees of a lime of 27°.
Ex. 2. Find tlie number of square ioches in the urea of a lune oE
27", on a spliere whose radius is 10 in.
A solid symmetrical with respect to a plane is a solid in which
a line drawn from any point in ila surface X the given plane and
produced its own length ends in a point ou the surface; hence
Ex. 3. How many planes of symmetry has a circular cylinder f A
cylinder of revolution !
Ez. 4. Haa eillier of these solids a center ot symmetry S
Ex. 5. Answer the same questions for a circular cone.
Ex. 6. For a cone of revolution. For a sphere.
Ek. 7. For a regular square pyramid. For a regular peiitBgonul
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458 BOOR IS. SOLID GEOMETET
pROPOsiTiox XXVI, Theorem
820. If two great circles intersect on a hemisphere, the
snm of two vertical triangles thus formed is equivalent to a
bme lohose angle is that angle in the triangles which is
formed by the intersection of the two great circles.
Given tbe liemispiiere ADBF, and on it the great circles
AFB and DFC, intcracetiiig at F.
To prove A AFC + A BFD ^ luue whose Z is BFV.
Proof. Complete the sphere aitd produce the given arcs
of the great circles to intersect at F' on the other hemisphere.
Then, in the A ^FCand BF'D,
araAF^RVcBF,
(each Uing the siipplemenl of the arc BF).
In like manner arc CF=are DF'.
And arc AC—ara J)B.
: A AFC~ A BF'B. Art. 79a.
Add the A BFD to each of these equals;
.-. A AFC + A BFD^A BF'D + A BFD. Ai, 3.
Or .-. A AF(7+A£FZ)o:luue fBFJ>. Ai. 6,
Q. £. I).
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SPHEBICAL AEEAS 459
Proposition XXVII. Theoreh
821. The niiinher of spherical dr-grees, or spherids, in
the area of a spherkal friaugle in equal to (he nnmber of
angular degrees in the spherical excess of the Irknigle.
Given the splierical A ABO whose A are denoted by ^,
B, C, and whose spherical excess is denoted by E.
To prove area of A ABC =i; spherids.
Proof. Produce the sides AC and BC to meet AB pro-
duced in the points D and F, respectively.
AXBC + A Gl>B = bnie ADDC = 2 A spherids. )
> Art. 816.
AABC+AACF=-hmQ£CI''A := 2 .0Brlieriils. )
A ABC + A CFP= luue of Z BOA ^ 2 C spherids. Art. >i2D-
Adding, and observing that A ABC+A CDJt+AACF
-I- A (7FD = hemisphere ABDFG,
2 A ABC+ hemisphere^? {A + B+ C) spherids. Or,
2 AAS(7+3COsphei-ids=2(.l + 7?+ C) spherids. Ax. s.
.-. A ABC + 180 spherids^ ( A +h+ C) splierids. Ax. 5.
.-. A ABC=(A + B+ C—-iiiO) spherids. Ax. 3.
Ur area A ABC=E spherids. Art. Tss.
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4(;0 ISOOK 1\. SOLID OKOlIETilV
822. Formiila for area of a spherical triangle m square
units of area.
Comparing the area o£ a spheri<'ul A with the area o£
the entire spliei-e ,
area A : 4 %]^ = E splierids : 720 spiierids.
4 7tR= X 7J . Tin-JE
:. area A = ^^ , or area A~— rr-r--
823. The spherical excess of a spherical polygon is the
sum of tlie angles of the polygon diminished by (h— 2) 180°;
that is, it is the sum of the spherical excesses of tiie tri-
angles into which the polygon may be divided.
Proposition XXVIII. Theorem
824. The area of a spherical polygon, in spherical de-
grees or spherlds, is equal to the spherical excess of the
Given a spherical polygon AHCDF of n aides, with its
spherical excess denoted by E.
To prove area of ABGDF^E spherical degrees.
Proof. Draw diagonals from A, any vertex of the poly-
gon, and thus divide the polygon into n—2 spherical A.
The area of each A= (sum of its i— 180) spherids.
Art. 821.
.■. sum of the areas of the A= [sum of /i of the ^ —
(«— 2) 180] spherids. As. 2.
.-. area of polygon = E spherids, Art. 823.
i/or the Diwi I'f A of the A— (.?i— 2) iau°=£°).
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SPHEIUCAL VOLUMES
SPHERICAL VOLUMES
825. A spherical pyramid is a por-
tion of a spliere bounded by a spheri-
cal polygon and the planes of the
great circles forming the sides of the
polygon. The base of a spherical
pyramid is the spherical polygon
bounding it, and the vertex of the
spherical pyramid is the center of tlie sphere.
Thus, in the spherical pyramid 0-ABCD, the base i
ABCD and the vertex is 0.
820. A spherical wedge (or ungula) is the portion of a
sphere hounded by a lune and the planes of the sides of
the lune.
827. A spherical sector is the portiou of a sphere gene-
rated by a sector of that semicircle whose rotation generates
the given sphere.
828. The base of a spherical sector is the zone gene-
rated by the revolution of the ai'c of the plane sector which
generates the spherical sector.
Let the pnpil draw a spherical sector in which the base
is a zouc of one buse.
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4t)l2 T!(K)K IX, WLH> (iEOMETKY
829. A Spherical segment is a portion of a sphere
ineluded between two parallel planes.
The bases of a spherical segment are the sections of the
sphere made by the parallel planes whlt:h bound the given
segment; the altitude is the perpendicular distance between
the bases.
830. A spherical segment of one base is a spherical seg-
ment one of whose bounding planes is tangent to the
sphere.
Pkoposition' XXTX. Theorem
831. The voluuu of a -ipheif is equal to oic-ihird the
product of the area of its suifiice hy its radiU"-
Given a sphere having its volume denoted by F, sur-
face by 5, and radius by R.
To prove 7=iSXK.
Proof. Let any polyhedron be eireurascribed about the
sphere.
Pass a plane through each edge of the polyhedron and
the center of the sphere.
These planes will divide the polyhedron into as many
pyramids as the polyhedron has faces, each pyramid hav-
ing a face of the polyhedn)H for its base, the center of the
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SPHERICAL VOLUMES 4R^
Sphere for its vertex, siiid the ra<lius of the sphert^ for its
altitude.
.'. volume of each pj'ramid = i base X R. (Why ?)
,'. volume of polyhedron = 4 (surface of polyhedron) XK.
If the number of faces of the polyhedron be increased
indefinitely, the volume of the polyhedron approaches the
volume of the sphere as a limit, and the surface of the
polyhedron approaches the surface of the sphere as a
limit.
Hence the volume of the polyhedron and h (surface of
the polyhedron) X R, are two variables always equal.
Hence their limits are equal.
Or "F-i«Xfi. (Why?)
832. Formulas for volume of a sphere. Substituting
^-4 7IE^ or S = 7il>- in the result of Art. 831,
v= — r— ; also 7=-— --
S8S. Cor. 1. The volumes of two spheres are to each
other as the cubes of their radii, or as the cubes of their
diameters.
834. CoK. 2. The volume of a spherical pyramid is
"/H'tl to one-third the product of its base by the radius of
thr .yyhere.
835. COK. 3. Th<' roliioie of spherical ^e'ior is equal to
■.<„>--lhi,-d the prodKcl of (7.S- haxe (the bomdiiuj zone) by the
i-<idnis of (he sphere.
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4G4
SOLID GEOJIETRY
836. Formula for the volume of a spherical sector. De-
noting the altitude of the sector by R ami the volume by T,
F= J ( area of zone ) X R,
= J (2 7iRH) n. Art. 8U.
PROrosiTiox XXX. Theorem
837. The volume of a spherical segment is equal to one-
half the product of its altitude by the sum of the areas of
its bases, plus the coliime of a sphere whose diameter is the
attitude of tli£ segment.
Given the semieirele ABCA' which generates a sphere
by its rotation about the diameter AA'; BD and CF semi-
chords X AA', and denoted by r and r'; and I>F denoted
by-ff.
To prove volume of spherical segmeut generated by
BCFD, or r=h{7ir^ + 7tr'-) R-{- \ TtiP.
Proof. Draw the radii OB and OC.
Denote Of by h, and OD by k.
Then 7=vol. 0B(7+ vol. OOP— vol. OBB.
:. y=^ Ttlt-U+^h Ttr'-h—h nr-k. Arts. 830, 723.
But U=h--li, A^=i^^r'^ andF^B-/--. (Whyf)
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SPHEIilCAL VOLUMES 465
= i 7t [2 Sf {li-k)-i (B-S') *— {B"-ri (■]. Ai. a.
= i Jt [2 E? (;i-f)+ H? (b-D-Ui'-fn-
= iKj[3ff-(P+M+F)].
4'— 2 Ji+F-iP. A-i. 4.
^3 B-|(,.= +, .»)-?."
r-j jiH [l (
.-. r-i (rtv=+jt.-=)ir+S nH'.
Q. E. B.
838. Formula for volume of a spherical segment of one
base. lu a spherical segment of one base r'~o, and )'"- =
(2K-ff)ir(Art. 343).
Substituting for r and r' these values in the result of
Art. 837,
'■-"'(^-t)-
839. Advantage of measurement formulas. The student
should observe carefully that, by the results obtained in
Book IX, the iceasurenieut of the areas of certain curved
sui-faees is reduced to the far simpler work of the measure-
ment of the lengths of one or more straight lines; in like
manner the measurement of certain volumes bounded by a
curved surface is reduced to the simpler work of linear
measurements. A similar remark applies to the results of
Book VIII.
Ex, 1. Find llip volume of a spheve whoso radius is T in.
Ex. 3. Find tlia voluiHe of a sphere wliose diompt^r is 7 in.
Ex. 3. In a spliore wIiohb radius is 8 in., find the volume of a
spheriful segmeiit of one base wliosu aUitiide is 3.
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466 HOOK TX. fiOT.TD r.F.OMETEY
EXERCISES. CROUP 72
THEOREMS CONCRRNlXr. THE SI'lfEliE
Ex. i. Of circles o£ a sphere whose pLmes pass thronph a civpn.
point within a sphere, tho smallest is thnt circle whoso plai.s ia
perpondieAilar to the diameter through tlio given point.
Ex. 2. If a point on tiie surface of a Riven sphere is equldiRtant
from three points on a given small circle of the sphere, it is the pole
of the small circle.
Ex. 3, It two sides of a spherical triangle are quadrants, the third
side measures the angie opposite that side in the triangle.
Ex. 4. If a spherical triangle has one right angle, the sum of its
other two augles is greater than One right angie.
Ex. 6. Thfl polar triangle of a biroctangular triangle is hirectan-
gular.
Ex, 7. The polar triangle of a trireotangular triangle is ideatieal
with the original triangle.
Ek, 8. Prove that the sum of the angles of a spherical quadri-
lateral is greater than 4 right angles, and less than 8 right angles.
What, also, are the limits of the sum of the angles of a spherical
hexagon I Of the sum o! the angles of a spherical n-gon ?
Ex. 9. On the same sphere, or on equal spheres, two birectan-
gular triangles are equal if tlieir oblique angles are equal.
Ex. 1 1. If one of the legs of a right spherical triangle is greate
than a quadrant, another side is also greater than a qiiaiJr.ant.
[Sua. Of the leg which is greater than a quadrant, take the em
remote from the right angle as a pole, and describe an arc]
Ex. 12. II ABC and A'B'C are polar triangles, the radius OA i
perpendicular to the plane OB'C.
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EXERCISES ON THE SPHERE 4b7
Ex. 13. On tlie same Ephere, or on equal sphecee, spherical tri-
angles whoae polar triangles have equal perimeters are equivalent.
Ex. 14. Given OAO', OISO', and All arcs of great
circles, interseoting so that Z OJC= Z O'BA ; prove that
^OAB=AO'AI!.
[SuG. Show that ZOB^I= ZO'JB.]
Ex. 15. Find the ratio o£ the volume of a sphere to
the volume oE a circumscribed cuho,
Ex. 16. Find the ratio of the surface of a sphere to
the laterr.l surtaee of a ciroumscribod cylinder of revolution; a'
find the ratio of their volumes.
Ex. 17. If the edge of a regular tetrahedron is denoted by a, fi
the ratio of the volumes of the inscribed and circumscribed ephe^j
Ex. 18. Find the ratio of the two segments into which a hen
sphere is divided by a plane parallel to the base o£ the UemispliBre b
at the distance jB from the ba-se.
EXERCISES. CROUP 73
Sl'HEKlCAL LOCI
Ex. 1. Find the locus o£ a point at a given distanne n from the
Bnrface of a given sphere.
Ex. 2. Find the locus of a point on the surface of a sphere that is
equidistant from two given points on the surface.
Ex. 3. If, through a given point outside a given sphere, tangent
planes to the sphere are passed, Snd the locus of the points ot
tangencj.
Ex. 4. If straight lines be passed through a given Ssod point in
space, and through anotiier given point other straight lines be passed
perpecdienlar to the first set, ilud the locus of the Xeet oi the
perpendiouiuia.
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4t)0 BOOK IX. SOLID GEOMETRY
EXERCISES. CROUP 74
rii015LT:MS COXCl':KMXfi THE STHEliB
El. 1. At a given point on a sphere, eonatruet a piano tangent to
the sphere.
Ex. 2. Through a given point on the siirfnco of a epliere, draw an
uro of s, great oirele perpendioulac to a givsn are.
Ex.3. Insoribe a circle in a given spherienl triangle.
Ex. 4. Construct a spherical triangle, given its polar triangle.
Given the radius, r, construct a spherical surface which shall pagg
through
Ex. 5. Three given points.
Ex. 6. Two given points and he tangent to a given plane.
Ex. 7. Two given points and be tangent to a given sphere
Ex. 8. One given point and be tangent to two given planes.
Ex. 9. One given point and he tangeut to two given spheres.
Given the radius, r, constrni^t a spherical siirfacQ which shall be
tangent to
Ex. 10. Three given planes,
Ex. 11. Two given planes and <jne given sphere.
Ex. 12. Construct a spherical surface whiah shall pass through
three given points and be tangent to a given plane.
Ex. 13. Through a given straight line pass a plane tangent to a
given sphere.
[Suo. Through the center of the sphere pass a plane J. [tiven
When is the solution iinpossiiule ?
Ex. 14. Through a given point on a sphere, construct an aru of
a great oirele tangent to a given small circle of the sphere.
[Sue, Draw a straight line from the center of the sphere to the
given point, and produce it to intevseot the plane of the small oirele,
etc.]
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NUMERICAL EXERCISES IN SOLID
GEOMETRY
For methods of facilitating immerical computations,
see Arts. 4i)3-6.
eXEROISES. CROUP' 7S
LINES AND SURFACES OP POLYHEDRON'S
Find the lateral area and total area oE a riglit prism whose
Ex. 1. Ease is au. equliatoral triaiigly of edge 4 in., aud whose
altitude is 15 in.
Es. 2, Base is a triangle of aides 17, 12, 25, aud whose altitude
is 20,
Ex. 3. Base is an isosceles trapezoid, the parallel bases being 10
and 15 and leg 8, and whose altitude is 24.
Ex. 4. Base is a rhomhus whose diagonals are 12 and 16, and
■whose altitude is 12.
Ex. 5. Base is a regular hexagon wUh side S tt., ar.i whose alti-
tude is 20 ft.
Ex. 6. Find the entire surface of a reetanguiar parallelepiped
8Xl2XlGiii.; ofonepX'jXrft.
Ex, 7. Of a eube whose edge is 1 ft, 3 in.
Ex. 8. The lateral area of a regular hexagonal prism is 120 sq. ft.
and an ed^-e o£ the base is 10 tt. Find the altitude.
Ex. 9. How many square feet of tin are neeessary to line a box
20X6X4in.?
Ex. 10. 1( the surfac'e of a c^ube is 1 pq.vd,, find an edge in inehes
E!t, 11. Find the diagonal of a cuhti whoso edge is 5 in.
Ex. 12. I£ the diagonal of a cube is 12 ft., find the surface.
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470 HOLID GEOMF.TliY
Ex. 13. If tho surfaop oEnreowngiilaipnrallolopipedia 208aq. in.,
and the odg03 are aa 2 ; y : 4, liud tUu oJjskh.
In a regular square pyramid
Es. 14. IE an udge o£ the base ia Id and eUut height is 17, find
the ftltitude.
nteral edge is 17, find nn edge
Ex. 16. If i\ lateral edge is 2:i aud an edge of the Ijiae is 14, find
the altitude,
In a regular ti'iangular pyramid
Ex. 17. If an edge of the baso is 3 and the altitude ia 10, find tho
slant height.
Ex. 18. Find the altitude of a regulav tetrabudron whouo edge \i 6,
Find the lateral surface and the total surface o£
Eic. 19. A regular square pyramid an edga ot wlioso base is 16,
and whose altitude is 15.
Ex. 22. A regular square pyramid whose slant height is 24, and
ihose laterai edge is 25.
Ez. 23. A regular tetrahedron whose edge ia 4.
Ex. 24, A regular tetrahedron whose altitude ia 9,
Ex. 26. In the frustum of a regular square pyramid the edges of
the bases are 6 and 18, and the altitude ia 8. Find the slant height.
Hence find the lateral area.
Ei, 27. In tho fru.^tum of a regular triangular pyramid the edges
ot the bases are i and 6, and the altitud. ia 5, Find the slant height.
Hence find the lattirul a-ttat.
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NUMERICAL EXERCISER IN SOLID GEOMETRY 471
Ex. 28. In tlie frustum of a regular tetrahedron, if the edge of the
lower base is hi, the edge of the upper base is 63, and the altitude is a,
show that L = il/i{b, — b-i)' +ia'.
Ex. 29. lu the fruatum of a regular square pyramid the edges o£
the bases are 2U and 60, and a lateral edge ia 101, Find the lateral
EXERCISES. CROUP ?e
LINES AND SURFACES OP CONES AND CYLINDERS
Ex. 1. How many square feet of lateral surface has a tunnel 100
I. long and 7 ft. in diaiueteiv
In a cylinder of revolution
Ex. 4. Find E in terms of S and H.
Es. 6. Find H in terms of E and T.
Ex. 6. Find T in terms of 5 and il.
Ex. 7. How many sq, yds of canvas are required to make a coni-
oal tent 20 ft. in diameter and 12 ft. liigb !
El. 8. A man has 400 aq. yds. of canvas and wants to make a
conical tent 20 yds. In diameter. What will be its altitude 1
Ex. 9. The altitude of a cone of revolution ia 10 ft. and the lat-
eral area is 11 times the area of the base. Find the radius of the baae.
In a cone of reTOlution
Ex. 10. Find T In terms of S and L.
Ex. 1 1. Find E in terms of T and L.
Ex. 12. How many square feet c
funnel the diametere of whoso ends
altitude ia 7 in. ?
Ex. 13. If the slant height of a fnistnm of a eoue of
makes an angle of 45" with the base, show that the lateral a
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SOLID GEOMETRY
EXERCISES, CROUP ^T
RPHr;R!CAT- LINES AND SL'RrACi:^
El, 2, How many squaro inelies of leather will It take to eovar a
bafieball whose diameter is 3^ in.f
Ex. 3. How many aq. ft. of tin are required to covpr a dome in
the shape of a liemiaphere 6 yds. in diameter ?
Ex. 4. What ia the radius of a sphere whose surface is filG sq. in. f
Ex. 5. Find the diameter of a globe whose surface is 1 sq. yd.
Ei. 7. If a hamispherical dome is to contain 100 sq. yds. of sur-
face, what must its diameter be V
Ex. S. Find the radius of a sphere in which the area of (he sur-
face equals the number o£ linear units in the circumference of a great
Find the area of a !une in which
Ex, 9. The angle of the lune is 3(i°, and the radius of the sphere
is 14 in.
Ex. 10. Tlie angle of the lune is lfi° 'M, and the diameter of the
Ex. 11. The angle of the lune is 24°, and the surface of the
sphere is i sq. ft.
Find the area of a spherical triangle in which
Ex. 12, Tiie angles are 80^, 90", 120°, and the diameter of the
sphere ia 14 ft.
Es. 13. The angles are 74° 24', 83° 16', 92° 20', and the radius of
the sphere is 10.
Ei. 14. The angles are 85°, S5°, 135°, and the surface of tlie
epliere ia 10 aq. ft.
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NUMERICAL EXERCISES IN SOLID GEOMETEY 473
Ex. 15. If the sides of n spherical trianKle nre 100°, 110^, 120"
and the radius of the sphero U 16, find the area of the polar triangle,
Ex. 16. It the angles of a spherical triangle are 00", 100". 1L>0'*
and its area is Bi)00, fiud tho radius of the sphere.
Ex. 17. If the area of an equilateral apherioai triangle is one-
third the surface of the spliere, fled an angle of the triangle.
Ex. 18. In a trihedral angle the plane angles ot the dihedral
angles are 80", i>0°, 100°; find the number of solid degrees in the
trihedral angle,
Ex. 19. Find the area of a spherical hexagon eanh of whoaa
Ex. 20. If each dihedral angle of a given pentahedral angle is
120", how many solid degrees does the pentahedral angle contain 1
Ex. 21. In a sphere whose radius is 14 in,, fiud the area of a zono
3 in, high.
Ex. 22. What is the area of the north temperate zono, if the
earth is taken to be a sphere vrith a radius of 4,000 miles, and tlie
distance between the plane ot the arctic circle and that of the tropic
of Cancer is 1,800 miles t
Ex, 23. If Cairo, Egypt, is in latitude 30°, show that its parallel
ot latitudft bisects the surface of the northern hemisphere.
Ex. 25. How much of the earth's surface will a man see who is
,000 miles above the surface, if the diameter is taken as 8,000 milesS
* of a zone equals the area of a groat circle,
zone in terms of the radius of the sphere,
Ex. 27. If sounds from the Krakato a explosion were heard at a
distance of 3,000 miles (taken as a chord) on the surface of the earth,
over what fraction of the earth's surface were they heard 1
Ex. 28. The radii of two spheres are 5 and 12 in. and their cen-
ters are 13 in, apart. Find the area of the circle of intersection and
also of that part of the surface of each sphere not Included by the
otiier ephere.
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474 SOLID GEOMKTKY
EXERCISES. CROUE' 7a
VOLUMES OF POI.YHEDKONS
Find the volume of a priam
Ex. 1. Whose base is an equilateral triangle ivitli Hide 5 !n., and
1j, and whose
Ex. 3. Whose bnae is an isoaeelea rigbt triangle with a leg equal
to 2 yds., and whose altitude is 25 ft.
Ex. 4. Whose base is a regular hexagon wvitli a side of 8 ft , and
whose altitude is 10 yds.
Ex. 5. Whose base is a rhombus one of whose sides is 2ii, and one
of whose diagonals is 14, and whose altitude is 11.
Ex. 6. Whose base contains 84 bc;. yds., and whose lateral fanes
are three rectangles with areas of 100, 170, 210 sq. yds., respectively.
Ex. 7. How many bushels of wheat are held by a bin 30x10x6 ft.,
if a tii^hel is taken as li cu. ft.?
Ex. 8. How many eart-loads of earth are in a collar 30 s 20 x G ft,
if a uart-load is a cubic yard ?
Ex. 9. If a etibioftl block of marble coats $3, what is the cost of
a tube whose edge is a diagonal of the first blocl; t
Ex. II. Find the edge of a cube whose volume equals the areii of
its surface.
Ex. 12. If the top of a cistern is a rectangle 12 s 8 £t., how deep
must the cistern be to hold 10,000 gallons F
Ex. 13. Find the inner edge of a peck measure which is in the
shape of a cube.
Ex. 14. A peck Taeasure is to be a rectangular patallelopipod with
square base and altitude equal to twioe the edge of the base. Find
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NUMEKICAL EXEHCISES IN SOLID GEOMETRY 475
Ex. IS. Find the volume of a cube whose diagonal is a.
Find the rolurae of a pyramid
Ex. 16. Whose base is an equilateral triangle with side 8 in,,
and whose altitude is 13 in.
Ex. 17.
one leg 21,
Whose
and wh
1 base is a right 1
ose altitude ie 20,
:rianele
with
hypotei
™.
>e 29 an
,d
Efc. 18. Whose base
lateral edges is 5.
is a square
with i
iide (>,
and ea.
?h
of who'
.e
Ex. 19. Whose
lateral fa^es mal^e
. hase
is a square
ngle of 45° V
with siiJe 10,
tith the base.
and en
ith
of whoi
,e
Ex 20
H the pvca'
re b^ise of si.
tain r What
mid of Memphis ha
ae 2:i;!yd3., how m
is this worth at $1 b
IS an altitude
■any cubic ya
>cu. yd.?
of
rds
146 yd
of stoi
Ex. 21. A ehiirch spire 150 ft. high is hesagonal in shape and
each side of the base is 10 ft. The spire has a hollow hexagonal
interior, eanh side of whose base is 6 ft., and whose altitude is 45 ft.
How many oubic yards of stone does the spire contain 1
I. yds. and its base is a square
Ex. 23. A heap of candy in the shape of a frustum of a regular
square pyramid has the edges of its bases 2ri and !] in . and its altitude
12 in. Find the number of pounds in the heap if a pound is a rectan-
gular parallelepiped 4x3x2 in. in size.
Ex. 24. Find the yoluma of a frustum of a regular triangular
pyramid, the edges of ths bases being 2 and 8, and the slant height 12.
Ex. 25. The edgea of the bases of the frustum of a regular square
pyramid are 24 and G, and each lateral edge is 13 ; Snd the volume.
Ex. 26. If a stick of timber is in the shape of a frustum of a
regular square pyramid with the edges of its ends and 15 in., and
with a length of 14 ft., End the number of feet of lumber in the stick.
What is the difference between this volume and that of a stick of
the sumo length having the shape of a prism with a base equal to tie
area o! a midsection ot the first sUck ;
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Ex.. 28. How many uavt-loada o£ eartli are there in n railroad cut
12 ft. deep, n-liose base is a rectangle 100 x 8 £t., and whose top is a
reetangle 30 x 50 It, I
Ex. 29. Find the volume ot a prismatoid whoso baae is an equi-
lateral triangle with aide 12 ft., nnd whose top is a Una 12 ft. long
liarallel to one Bide of the base, and whose altitude is 15 ft.
Ex, 30. If the haae of a prismatoid is a rectangle with dimensions
a and b, the top is a line c parallel to the aide !i of the liase, and the
altitude ia 'i, find the vulume.
EXERCISES. CROUP ta
VOLTTJIES OF COXES AND CYLlNDii;RS
Ex. 1. How many barrels of oil are contained in a cjlicdrical
tank 20 ft. long and G ft. in diameter, if a barrel contains 4 cu. ft. t
Ex. 2. How many ou. yds. of earth must be removed in making a
tunnel 450 ft. long, it a eross-seotion of the tunnel is a aeiEiicirole of
15 ft. radius f
Ex. 3. A cylindrical glass 3 in. in diameter holds half a pint.
Find its height in inches.
Ex. 4. If a cubic foot of
lameter, how long will the w
altitude equals the
Ex. 6. Showthat the YOlumesof two cylinders, having the lUtitude
f each equal to the radius of the other, are to each other as E : S'.
Ex. 8. A conical heap of potatoes is 44 ft. in circumfeienee and
6 ft. high. How many bushels does it contain, if a bushel is lieu. It.t
e- glass hold, it
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NTTMEEICAL EXERCISES IN SOLID GEOMETItY 477
Ex. 10. Find the ratio of tho volumea o{ the two eonts ins^iribed
ribed about, a regular tetrahedron.
X. 11. IE an equilateral eono eontaius 1 quart, find its diraen-
b of R aud L ; alao
Ejc. 13. Find tiie volume of a frustum of a cone of revolution,
whose radii are 14 and T ft., and whose altitude is 3 yds.
Ex. 14. What is the cost, at 50 cts. a oh, ft., of a piece of marble
in the shape of a fruBtum of ft cone of revolution, whose radii are 6
and 9 ft., and whose slant height is 5 ft.?
ExenciSEs. crouf so
SPHERICAL VOLUJLES
Ex. 1. Find the volume of a sphere whose radius is 1 ft. 9 in.
Ex. 2. I! the earth is a sphere 7,9L'0 miles In diameter, Bnd iti
Ex. 3. Find the diameter of a sphere whose volume is 1 cu. ft.
Ex. 4. What is the volume of a sphere whose surface isClG sq. in.'
Ex. 5. Find the radius of a sphere equivalent to the sum of twi
pheres, whose radii are 2 and 4 in.
Ex. 7, Find the volume of a sphere oircumseribed
whose edge is fi.
Ex. 8. Find the voluiue of a spheiifial shell whc
outer diameters are 14 and 21 iu.
Ex. 9. Find the volume o( u spberii.'al shell whoso ii
surfaces :izo 20 tt wnd i: tt.
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4 1 8 SOLID GKOJIETllY
Ifind tiio voluBid of
Es. 10. A splierieal wedge whoso auglo is 3!", the radius of the
sphere being 10 in.
Ex. 11. A spherical Keetor whoso base is a. zone 2 in. high, the
rartius of the sphere being 10 in.
a whose rsdii are 4 and
Ex. 13. A wash-basin in the shape r>£ a aogment, of a sphere is
6 in. deep and 2i in. in djonieter. How many quarts o£ water will the
basin hold t
Ex. 14. A plana parallel to the base of a hemisphere and bisect-
ing thi! altitude diifides its volume in what ratio f
Ex. 15. A aplieri<^al Kogmeut 4 in. high «oiitains 200 eu. in. ; End
the r.idiu.'i of the sphere.
Ex.16. If a heavy sphere whoso dia
ponieal wice-Glass full of water, whose i
6 in., find how much water will run over
EXERCISES. GROUP 81
EQUIVALENT SOLIDS
Ex. i. IE a cubical block of putty, eaffh edge ot whi
ch is 8 incheg,
be molded into a cylinder oE revolution whose radius h
1 3 inches, find
the altitude ot the eyiicder.
Ex. 2. Find the radius of a sphere equivalent to
a cube whose
edge ie 10 in.
Ex. S. Find the radius of a sphere equivalent to a i
;one of revolu-
tion, whose radius ia 3 in. and altitude G in.
Ex. 4. Find the edge of a cube equivalent to a frustum of a eons
f revolution, whose radii are 4 and 9 ft. and altitude 2 yds.
Ex. 5. Find the altitude of a rectangular parallelopiped, whose
ase is 3x3 in. and whose volume Is eqaivaleot to a spheru of tadiua
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NUMERICAL EXERCISES IN SOLID GEOMETlil' 479
Ex. 6. Find the biiRe of a square reetangular pai'allelopiped,
Vfliose altitude ia 8 in. aud whose volume equals the volume o£ a cone
of revolution with a radiuB of 6 and au aititudo of 12 m.
Ex. 7. rind the radius of a cone of revolution, whose altitude is
15 and whose volume is equal to that of a eyliniier of revolutiou ivith
radius G and altitude 20.
Ex. 8. Find the altitude of a eone of revolution, whoso radius is
15 and whose voiume equals the volume of n cone of revolulion with
radius 9 and altitude 24.
Ex, 9. On a sphere whose diameter is 14 the altitude of a zone of
one base is 2. Find the altitude of a pylinder of revolution, whose
base equals the base of the zone and whose lateral surface equals the
surface of the zone.
EXERCISES. CROUF*
SIMILAE SOLIDS
Ex. 1. If on two similar solids L, L' and !, I' are
ORous lines; -4, A' and a, «' pairs of homologous
T, v' pairs of homologous volumes,
pairs of homol-
veas, V, V and
ft :L' = 1 :I' = v'-^ ■.i/A' = ^'v.f'e'.
show that - A : A' = L^ : L" = a : n'= r* : FL
Ex, 2. If the edge of a cube is :o in., &nd the
having 5 times the surface.
edge of a cube
Ex. 3, If the radius of a sphere is 1(1 in., find the radiu
Bphere having 5 times the surface.
Ex. 5. In the last three exercises, find the required di
10 volume ia to be 3 times the volume of the original solid.
those of auotliui- trunk. How
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SOTJD GEOMliTUy
Ex. 7. How fill
from Iho vertex is
i Ihe croBK-section which bisects
tho volume of a
cone Ot I'BVOlutiOI
1 f Wliich biseeta tiie lateral
Eurfaee ?
Ex. 8. If the nltiliide oC a pyramid is bisected Tjy a plane parallel
to the 1)sse, how does the area of tho cross-seetion ootupare with the
area of the liase ? How does the volume cut olf compare with the
volume of the entire pyrnmid ?
Ex. 9. Plnnes parallel to Iho base of a coae divide tho altitude
into three equal pai'ts ; compare the lateral surfaces cut off. Also the
volumes'.
Ex. 10. A sphere 10 i(i. in diameter ia divided into three equiva-
lent parts by concentric spherical surfaces. Find the diameters of
toese surfaces.
Ex. 11. It the strength of a muscle is as the area of its cross-
section, and Goliath of Gath was three times as large ia each linear
dimension as Tom Thumb, how much gre;iter was his strength ! His
weight f How, then, does the activity of the one man compare with
that of the other ?
Ex. 12. If the rate at which heat radiates from a body is in pro-
portion to the amount of surface, and the planet Jupiter has a diame-
ter 11 times that of the earth, how many times longer will Jupiter be
in cooling off t
[SuG. How many times greater is the volume, and therefore the
original amount of heat in Jupiter ? How many times greater is its
surface f What will be the combined effect of these factors T]
CROUP 83
MISCELLANEOUS NUMERICAL EXERCISES IN SOLID
GEOMETRY
Find S, T and V of
Ex. 1. A right triangular prism whose altitude is 1 ft., and t!
Bides of whose base are 26, 28, 30 in.
base ia 1ft. 2ii
Bx. 3. A frustum of a square pyramid Iho areas of whose has
are 1 sq. It, and 3G sq. in,, and whose altitude ia !) ia.
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NUMERICAL EXEECISES IN SOLID GEOMETRY 48X
Ex. 4. A pyramid whose slant height is 10 in., and whose base
I an equilateral triangle whose side is 8 in.
Ex. 6. A frustum o( a cooe of revolutioo whose radii
11 in. and slant height 13 in.
Ex. 8. Find the volume of a sphere inscribed in a cube whose
edge is 6; also find the area of a triangle on that sphere whose anglea
are 80°, 90", 150°.
> of the spherical pyramid whose base is
Ex. 10. Find the angle of a lime on the same sphere, equivalent
to that triangle.
Ex. II. On a cube whose edge is 4, planes through the midpoints
of the edges eut oE the corners. Find the volume of the solid re-
maining.
Ex. 12. How is F ehnnged if n of a cono of revolution is doubled
and R remains unchanged t If R is doubled and W remains unchangedt
If both H and J! are doubled ?
Ex. 13. In an equilateral cone, find S and V in terms of E.
Ex. 14. A piece of lead 30 x H :t 3 in. will make how many spher-
ical bullets, each S in. in diameter !
Ex, 16. How many bricks are necessary to make a chimney in
the shape of a frustum of a eone, whose altitude is DO ft., whose outer
diameters are 3 and 8 ft,, and whose inner diameters are 2 and 4 ft.,
eonnting 12 bricks to the eubio ft. ?
a of a zone is 300 and its altitude 6, find t
Ex. 18. If every edge of a square p7iamidisit Qtid^in termsof T.
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453 SOLID GKOMliri'llI:
Ex, 19, Arpgiikr ^qHarc pyramid hns a for its allituJe and also t ,,.
encli side of the liiu-e. Find llie area of a section made by a plane parallel
to the base mid bifieGting Ilia altitude. Find also tlie volumea of tha
two parts into wliieli the pyramid ia divided.
Ex. 20. It the earth is a sphere of 8,000 milos diameter and its
alroosphere extends 50 miles from the eavlb, find the volume of the
Ex. 21. On a spherp, find the ratio of llie area or an e.iuilateral
ipberipal Iriiinglp. eaeh of wliose angles ia 35', to liie area of a lune
Ex. 22. A square right prism has an altitude Cm and an edge o£
the base 2a. Find the volume of the largest cylinder, sphere, pyramid
ami cone which can be ent from it.
Ex. 23. Obtain a formula for the area of that part of a'sphere
illujuinateit by a jioint o£ light at a distance a from the sphere
whose radins is H.
Ex. 24. On a sphere whose radius is 6 in., find an angle of an
equilateral tnaBjtle whose area is 12 sq. in.
Ex. 25. Find the volume of a priemiitoid, whose altitude is 2-1 and
whose bases are equilateral triangles, each side 10, so placed that tlie
raid-seotion of the prismatoid is a regular hexagon.
Ex. 26. Ou a sphere tihose radius is 16, the bases of a zone are
equal and are together equal to the area of the zone. Find the alti-
tude of the zone.
e of a square pyramid, the edge of whose
teral edges is inclined G0° to the base.
Ex. 28. Aq irregular piece of ore, if placed in a cylinder partly
filled with water, causes the water to rise 6 in. If the radius of the
cylinder is 8 in., what is the volume of the ore ?
Ex. 29. Find the volume of a truneated right triangular prism, if
the edges of the base are 8, 9, 11, and the lateral edges are 12, 13, 14.
Ex. 30. In a sphere whose radius is 5, a aectioii is taken si tho
distance 3 from the center. On this section as a base a oone iu formoti
whose lateral elements are tangent to the sphere. Find tho lateral
surface and volume of the coua.
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NUMERICAL EXEECIRES IS SOLID GEOatETRY 483
Ex.31. The volume of a sphere 13 l,437iau. in. Find the surface.
Es. 32. A square whose side is 6 is revolved about a diagonal as
a axis ; find the surface and volume generated.
Ex. 33. Find the edge ot a cubical cistern that will hold 10 tons
f water, if 1 cu. ft. of water weighs 62.28 lbs,
Ex. 34. A water trough has equilateral triangles, eaeh side 3 ft.,
)r ends, and is IS ft. loiip. How many buckets of water will it hold,
! a bucket ta a cylinder 1 ft. in diameter and li ft. high J
Ex. 35. Tlie lateral ar.
ea of a (.-yiinder of r
evolution is 440 sq.
d the volume is 1,5411 cu
.. in. Find the ra.lli
IS and altitude.
Ex. 36. The anfrles of a spherical quadrilateral are 80", 100",
1-0", 120°. Find tile angle ot an equivalent equilateral triangle.
Ex.37. A cone and a cylinder have equal lateral suifacep, and
their axis sections are equilateral. Find the ratio of their volumes.
Ex.38. A water-pipe I in. 1
How many quarts of water mu
from under the ground oemes i
how long must the watei' run f
Ex. 39. A cube immersed
causes the water to rise 4 in.
what is an edge ot the cube 1
Ex. 40. An auger hole whos
Ex. 42. The volumes ot two similar cylinders of revolution are
as 8 : lliS; find the ratio of their radii. If the r.tdius of the smaller if
10 in., what is the radius of the larger?
Ex. 43. An iron shell is 3 in. thick and the di^imeter of Us oute:
surface is 28 in. Find its volume.
Ex. 44. The legs of an isosceles spherical triangle each make fir
".ngle of Ta° «ith the base. The legs [>i'odueed form a iiine whosi
aiea is four times the area of the triangle. Find the angle of the lune
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484 SOLID CEO^iirrsY
CROUP 81.
EXERCISES INVOLVING THE MISTRIC S^YSTEM
rind s. r, r o(
Ex. 1. A riKht priam tho edRes of nhoso bas^o are li m., 70 dm.,
900 cm., and wbose altitude is 90 dm.
Ex. 2. A regular square pyramid sn fidge of whose base is 30 lim.,
and whose altitude is 1.7 m,
Ex. 3. A sphere whose radius is 0.02 m.
Ex. 4. A frustum of a oone of reTolution whose radii are 10 dm
acd 6 dm., and whose slant height ia 50 cm.
Es. 5. A cube whose diagonal is 12 cm.
Ex. 6. A cylinder of revolution whose radius equals 2 dm., auii
■whose altitude equals the diameter of the base.
Ex. 7. Find the area of a sphericnl trianglp on a sphere who^f
radius is 0.02 m., if its angles are 110°, 120°, 130°.
Ex. 8. Find the number of square meters iu the surface of [
sphere, a great circle of which is 50 dm. long.
Ex. 9. How many liters will a cylindrical vessel hold that i
10 dm. in diameter and 0.iJ5 m. high ? How many liquid quarts ?
a cylinder whose diameter is half th.
Ex. 1 1. Find the surface of a sphere whose TOlume is 1 cu. m.
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APPET^DIX
I. MODERN GEOMETRIC CONCEPTS
840. Modern Geometry, In recent times many new
georaetfie ideas have been invented, and some of them
developed into important new branches of geometry.
Thus, the idea of symmetry (see Art. 484, etc.) is a
modern geometric concept. A few other of these modero
concepts and methods will he briefly mentioned, but
their thorough consideration lies beyond the seope of
this hook.
841. Projective Geometry. The idea of projections
(see Art. 345) has been developed in eomparatiyely
recent times into an important branch of mathematics with
many praetieal applications, as in engineering, architec-
ture, constructicm of maps, etc.
842. Principle of Continuity, By this principle two
or more theorems are made special cases of a single more
general theorem. An important aid in obtaining continuity
among geometric principles is the application of the con-
cept of negative quantity to geometric magnitudes.
Thus, a negative line is a line opposite in direction to
a given line taken as positive.
For example, if OA is +, OB is -
(485)
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48G
Similarly, a negative angle is
an angle formed by rotating a line
in a plane in a direction opposite
from a direction of rotation taken
as positive. Thus, if the line OA
rotating from the position OA
forms the positive angle AOB, the
same line rotating in the opposite direction forms the
negative angle AOB'. Similarly, positive and negative
ares are formed.
In like manner, if P and P' are
on opposite sides of the line AB and
the area PAB is taken as positive,
the area P'AB will be negative.
As an ilhistratiou of the law of
continnity, we may take the Uieoreni
that the sum of the triangles formed
by drawing lines from a point to the
vertices of a polygon equals the area
of the polygon, " "
Applying this to the qnadrilateral ABCD, if the point
Pfalla within the quadrilateral, ^PAB+AFBC + APCI)
+ i^rAD=ABCr> (Ax. 6).
Also, if the point falls without the quadrilateral at J",
AP'AB + AP'BG+ AP'C7>+ A P'AD == ABCD, since
APAD is a negative area, and hence is to be subtracted
from the sum of the other three triangles.
843. The Principle of Reciprocity, or Duality, is a
principle of relation between two theorems by which each
theorem is convertible into the other by causing the woi-ds
for the same two gooniotrie objects in each tlicorem to ex-
cliaTige plaecK.
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MODERN GEOMETRIC CONCEPTS 487
Thus, of t.liL'orems VI and VII, Book I, either may be
convei'ted into the otiier by replacing the word "sides" by
"aogles," and "angles" by "sides." Henee these are
termed reciprocal theorems.
The following are other instances of reciproeiil geometric
properties :
1. Two points
2. Three p<ri«U
straight line deleriN
determine a
<e a plane.
1. Tw
2. Tin
lines lie t ermine a point.
ee plaim 7inl through the
ight line ileierniine a
3. A straight U
AeUnnine a piune.
" <""' « I'O'"'
3. Js
deter iiiiiw
fraighl line wul a plane-
a i'oiiit.
The reciprocal of a theorem is not necessarily true.
Thns, two parallel straight lines determine a plane, but
two parallel planes do not determine a line.
However, by the use of the principle of reciprocity,
geometrical properties, not otherwise obvious, are fre-
quently suggested.
844. Principle of Homology. Just as thii law of reci-
procity indicates relations between one set of geometric
concepts (as lines) and another set of geometric concepts
(as points), so the law of homology indicates relations
between a set of geometric concepts and a set of concepts
outside of geometry: as a set of algebraic concepts, for
instance.
Thus, if a and l are nnmbers, by algebra {a + h) (a — i)
Also, if a and b are segments of a line, the rectangle
(«+6)X (i( — h) is equivalent to the difference between the
squares ir and V'.
By means of this principle, truths which would be over-
looked or difficult to prove in one departmeut of thonght
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488 (lEO^liriKY. Al'I'ENVlX
are made obvious by observing tlie corresponding truth in
another department of thought.
Thus, if a and h ai-e line segments, tlie theorem (a + h)"
+ {a—b}- — 2{(t--i-b-} is not immediately obvious in geo-
metry, but becomes so bs observing the Hke relation
between the algebraic uumbers a and i>.
845. Non-Euclidean Geometry. Hyperspace. By vary-
ing the properties of space, as these are ordiuarily stated,
different kinds of space may be conceived of, each having
its own geometric laws and properties, Thus, space, as
we ordinarily conceive it, has three dimensions, but it is
possible to conceive of space as having four or more
dimensions. To mention a single property of four dimen-
sional space, in sneh a space it would be possible, by
simple pressure, to turn a sphere, as an orange, inside
out without breaking its surface.
As an aid toward eonoeiving how tliia ia poaslblo, consider a plane
in wiiicil one eii'cle lies inside another. No matter how these eirelea
are moved about in the plane, it is impossiljle to shift the inner eirale
KO aa to piaee it outside the other witliout l)rea!iing the oiroumference
III the outer circle. But, if we are allowed to use the third dimension
of space, it ia a aimple matter to lift the inner circle up out of the
plane and set it down outside the larger oirale.'
Suuilarly it, in space o£ three dimensions, we have one spherical
aliell inside a larger shell, it is impossible to place the smaller shell
ontsida the larger without breaking the larger. But if the nse of a
fourth dimension be allowed, — that is. the use of another dimension
of freedom of motion, — it ls possible to place Che inner shell outside
the larger without breaking the latter.
846. Curved Spaces. By varying the geometric, axioms
of space (see Art. 47), different kinds of space may be
conceived of. Thus, we may conceive of space such that
through a given point one line may be drawn parallel to a
given ime (that is ordmary, or Euclideau space}; or such
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MODF.KS GEOMETillC COKCEPTS 489
that through a given point no line can be drawn parallel
to a given liue (spherical space); or such that through a
given point more than one line can be drawn parallel to a
given liue (pseudo- spherical space).
These different kinds of space differ in many of their
properties. For example, in the first of them the sum of
the angles of a triangle equals two right angles; in the
second, it is greater; in the third, it is less.
These different kinds of space, however, have many
properties in common. Thus, in all of them every point in
the perpendieaiar hiseetor of a liue is equidistant from the
extremities of the liue.
eXERCISKS. CROUP SE
Ex. 1. Show by the use of zero and negative ares that the princi-
ples of Arts. 257, 263, 258, 264, 203, are particular cases of the genei'al
theorem that the angle inoluded between two lines which cut or tooth
a cirule is measured hj oue-lialf the sum of the intercepted area.
Ex. 2, Show that the principles of Arts. 354 and 3o8 are particular
eases of the theorem that, if two lines are drawn from or through a
po'nt t m t a eireumfa ranee, the product o£ the segments o£ one line
equ h p duct of the segments of the other iino.
Ex 3 bhow by the use o£ negative angles
tha h em XXXVIII, Book I, is true for a
quad a e al o£ the form AISCD. IBGD is a
ne a e an le; the angle at the vertex i» is
the fl sa leADC]
Ex 4 What is the reciprocal of tiie state-
Ex. 5, What is
perpendicular to i
dicular to eiich oti
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IT. ITIRTORY OP GKO:\IETRY
847. Origia of Geometry as a Science. The beginumgR
of geometty as a science are found in Kgypt, diitiug back
fit least tliree thonsaud years before Christ. Herodotus
says that geometry, as known iu Egypt, grew out of the
need of remeasuring pieces of laud parts of which had been
washed away by the Xiie floods, in order to make an e^ni-
tabie readjustment of the taxes on the same.
The substance of the Egyptian geometry is found in an
old papyrus roll, now iu the British museum. This roll
is, in eSect, a mathematical treatise written by a scribe
named Ahmes at least 1700 B.C., and is, the writer states,
a copy of a more ancient work, dating, say, 3000 B. C.
848. Epochs In the Development of Geometry. Prom
Egypt a knowledge of geometry was transferred to Greece,
whence it spread to other countries. Hence we have the
following principal epochs in the development of geometry:
1. Egyptian : 3000 B. C— 1500 D. C.
2. Greek : COO B. C— 100 B, C.
3. Hindoo : 500 A. D— 1100 A. D.
4. Arab : 800 A. D.— 1200 A, D.
5. European : 1200 A. D,
In the year 1120 A. D.. Athelard, an English monk,
visited Cordova, m Spam, in the disguise of a Mohamme-
dan student, and proeui-cd a copy of Euclid m the Arabic
language. This book he brought back to central Europe,
where it was translated into Latin and became the basis of
all geometric study in Europe till the year lriJ3, when,
(WO)
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niKTORY OF GEOMETRY 4i'l
owing to the capture of Ooiiatautiiiople by the Turks,
copies of the worlis of the Greek mathematicians in the
original Greeli were scattered througli Europe.
HISTORY OF GEOMETRICAL METHODS
849. Rlietorical Methods. By rhetorical methods in
the presentation of geometric truths, is meant the use of
definitions, axioms, theorems, geoiiieti'ic figures, the rep-
resentation of geometi'ic magnitudes hy tlie use of letters,
tlie arrangement of material in Books, etc. The Egyptians
iiad none of these, their geometric knowledge being re-
corded onlj' in the shape of the solutions of certain numeri-
cal examples, from which the rules used must be inferred.
Thales (Greece COO B.C.) fii'st made an enunciation of
an abstract pro]»ei'ty of a gcometi'ie figure. He had a rude
idea of the geometric theorem.
Pythagoras (Italy 535, B.C) introduced formal defini-
tions into geometry, though some of those used by him
were not very accurate. Fof instance, his definition of a
point is "unity having position." Pythagoras also
arranged the leading propositions known to him in
something like logical order.
Hippocrates (Athens, 420 B. C.) was the tirst systemati-
cally to denote a point by a capital letter, and a segment
of a line by two capital tetters, as the line AB, as is done
at present. He also wrote the first text-book on geometry.
Plato (Athens, 380 B.C.) made definitions, axioms
and postulates the beginning and basis of geometry.
To Euclid (Alexandria, 280 B. C.) is due the division
of geometry into Books, the formal enunciation of tlico-
rems, the particular enuucialion, the f(>nual constru(;rion,
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492 GEOMETRY. Arl'ENDrX
proof, and cuuclusioii, Ui presenting a proposition. He
also introduced tlie use of the corollary and scholium.
Using these methods of presenting geometric truths,
Euclid ■wrote a text-book of geometry iu thirteen books,
which ■was the standard text-book on this subject for
nearly two thousand years.
The use of the symbols A, CO , \\ , etc., in geometric
proofs originated in the United States iu recent years.
850. Logical Methods. The Egyptians used no formal
methods of proof. They probably obtained their few crude
geometric processes as the result of experiment.
The Hindoos also used no formal proof. One of their
writers on geometry merely states a theorem, draws a
figure, and says "Behold I "
The use of logical methods of geometric proof is due to
tiie Greeks. The early Greek geometricians used experi-
viental methods at times, in order to obtain geometric
truths. For instance, they determined that the angles at
the base of an isosceles triangle are equal, by folding half
of the triangle over on the altitude as an axis and observ-
ing that the angles mentioned coincided as a fact, but
without showing that they must coincide.
Psrthagoras (i325 B. C.) was the first to establish geo-
metric truths by systematic deduetion, but his methods
were sometimes faulty. For instance, he believed that the
converse of a proposition is necessarily true.
Hippocrates (420 B. C.) used correct and rigorous de-
duciion iu geometric proofs. He also introduced specific
varieties of such deduction, such as the method of reduc-
ing one proposition to another (Art, 296) , and the reductio
ad ahsurdtim.
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HISTORY OF GSOMETKY 493
The methoda o£ deduction used by the Greeks, ho^vaver, were de-
fective in their lack of generality. For instance, it was often thought
necessary to have a separate proof of a theorem for each different
kind of figure to which the theorem applied.
Thus, the theorem that the sura of the an- y-~~~~7\
gles of a triangle equals two right angles ,-' 'v, /' '\
(1) for the equilateral triangle Tiv use '•■ / .. . . v
of the regular hexagon ;
(2) for the right triangle by the use of
a rectangle;
(3) tor a Boalene triangle by dividing
the scalene triangle into two right triangles.
The Greeks appeared to fear that a
general proof might be vitiated if it were
applied to a figure in any way special or
peculiar. In other words, they had no conception of the principle o(
eontinuity (Art. 843).
Plato (380 B. C,} introduced the metliod oE proof by
ait<'!ysi'^. that is, by taliing a proposition as true and work-
ing from it back to known truths (see Art. 196) .
To Eudozus (3G0B. C.) is virtually due proof by the
method of limits; though his inetliod, known as the
method of exhaustions, is crude and cumbersome.
ApoUonius (Alexandria, 225 B. C.) used projections,
transversals, etc., which, in modern times, have developed
into the subject of projective geometry.
851. Mechanical Methods, Tlie Greeks, in demonstra-
ting a geometrical theorem, usually drew the figure em-
ployed in a bed of sand. This method had certain advan-
tages, but was not adapted to the use of a large audience.
At the time when geometry was being developed in Greece, the
interest in the subject was very general. There was scarcely a town
but had its lectures on the subject. The news of the discovery ol a
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4',)i GEOJIETKY. .VPI'EyDIX
new theortm spreati from town to town, and the theorem nas redenjon-
strated in the sanii of each raarket place.
The (ireek treatises, however, were written on velhim
or papyrus by tiie use of the reed, or calamus, and ink.
In Roman times, and in the middle ages, geometrienl
figures were drawn in wax smeared on wooden boards,
called tablets. They were drawn by the nse of the stylus,
a metal stieii, pointed at one end for making marks, and
broad at the other for erasing marks. Tliese was tablets
were stiil in nse in Shakespeare's time (see Hamlet Act 1,
Sc. 5, 1, 107). The blackboard and crayon are modern
inventions, their nse having developed within the last one
hundred years.
The Greeks invented many kinds of drawing instruments
for tracing various curves. It was due to the inflneuee of
Plato (1180 B. C.) that, in constructing geometric figures,
the use of oniy the rnier and compasses is permitted.
HISTORY OF GEOMETRIC TRUTHS. PLANE GEOMETRY,
852. Rectilinear Figures. The Egyp-
tians measured the area of any fonr-
sided field by multiplying half the sura
of one pair of opposite sides by half the
sum of the other pair; which was equivalent to using the
lormuia, area — — ^X -n ~'
This, of ooursB, gives a eorreet result loc the rectangle and square,
but gives too great a result for otlier quadrilaterals, as the trapeaoid,
etc. Hence Joseph, of the Book of Genesis, in buying the fields of
the Egyptians for Pharoah in time of famine by the use of this
(ormuia, ill many onses paid for a larger field than lie obtained*
The Egj-ptians had a special fondness for geometrical
constructions, probably growing out of their work as temple
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HISTORY OF GEOMETRY 495
biiilders. A ekss of workers existed among them called
"rope-stretchers," whose business was the marking out of
the foundations of buildings. These meu knew how to
bisect an angle and also to coustruct a right angle. The
latter was probably done bj' a method essentially the same
as forming a right triangle whose sides are three, four and
five units of length. Ahmes, in his treatise, has various
constructions of the isosceles trapezoid from different data,
Thales (600 B, C.) enunciated the following theorems:
If two straight lines intersect, the opposite or vertical
angles are equal;
The angles &i the base of an isosceles triangle are equal;
Two triangles are equal if two sides and the included
angle of one are equal to two sides and the included angle
of the other;
The sum of the angles of a triangle equals two right
angles;
Two mutuallj" equiangular triangles are similar.
Thales used the last of these tiieorems to measure the
height of the great pyramid by measuring the length of
the shadow cast by the pyramid and a!so measuring the
length of the shadow of a post of known height at the same
time and making a proportion between these quantities.
Pythagoras (525 B. C.) and his followers discovered
correct formulas for the areas of the principal rectilinear
figures, and also discovered the theorems that the areas of
similar polygons are as the squares of their homologous
sides, and that the square on the hypotenuse of a right
triangle equals the sum of the squares on the other two
sides. The latter is called the Pythagorean theorem.
They also discovered how to constnict a square equivalent
to a given parallelogram, and to divide a given line in
mean and extreme ratio,
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496 GEOMETRY. Ari'ENDIX
To EudoxuB (380 8. C.) we owe the general theory of
proportion in geonietr.v, and the treatment of incommen-
surable quantities by the method of Eshaustions. By the
use of these he obtained such theorems as that the areas
of two circles are to each other as the squares of their
radii, or of their diameters.
In the writings of Hero (Alexandria, 125 B. C.) we first
find the formula for the area of a triangle in terms of its
sides, K=Vs(s — a) (s — b) (s — e). Hero also was the first
to place land-snrveyiug on a scientific basis.
It is a curious fact tliat Haro at the same time gives an ineorreet
formula for the area of a triangle, ¥iz., K=ia{i+':), tliis formula
being apparent!}' derived from Egyptian sources.
Xenodorus (150 B. C.) investigated isoperemetricai
figures.
The Romans, though they excelled in engineering, ap-
parently did not appreciate the value of the Greek geom-
etry. Even after they became acquainted with it, they
continued to use antiquated and inaccurate formulas for
areas, some of them of obscure origin. Thus, they used
the Egyptian formula for the area of a quadrilateral,
K=—^ X— ^. They determined the area of an equilat-
eral triangle whose side is a, by different formulas, all
incorrect, as K=~~- , K=i{a"+a), and ZL = ia^.
853. The Circle. Thales enunciated the theorem that
every diameter bisects a circle, and proved the theorem
that an angle inscribed in a semicircle is a right angle.
To Hippocrates (420 B. C.) is due the discovery of
nearly all the other principal properties of the circle givaa
in this book.
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HISTORi" OP GEOMETRY 497
The Egyptians i-pgarded the area of the circle an eL|iii\ni-
lent to ,ff of the diameter squared, which would make
T-3.iC04.
The Jews and Babylonians treated 7t as equal to 3.
Archimedes, by the use of inscribed and circumscvihed
regular polygons, showed that the true value of 7t lies
between Sf and 3ki; that is, between 3.14285 and 3.1408.
The Hindoo writers assign various values to 7t, as 3, 3b,
l/lO, and Aryabhatta (530 A. D.) gives the correct ap-
proximation, 3.1416. The Hindoos iised the foi'mnla
1/2 1/4 A& '^^^ ■^''*- *^^^ ^^ computing the numeri-
eal value of ti.
Within recent times, the value of n has been computed
to 707 decimal places.
The use of the symhoi n for the ratio of the circum-
ference of a circle to the diameter was established in
mathematiea by Euler (Germany, 1750).
HISTORY OF GEOMETRIC TEDTHS, SOLID GEOMETRY
854. Polyhedrons. The Egyptians computed the vol-
umes of solid figures from the linear dimensions of such
figures. Thus, Ahmes computes the contents of an Egyp-
tian bani by methods which are equivalent to the use
is not known, it is not possible to say whetlier this formula
is correct or not.
Pjrthagoras discovered, or knew, all the regular poly-
hedrons except the dodecahedron. These polj-hedrons were
supposed to have various mugica! or mystical properties.
Heuce the study of them was made very iiromiaeut,
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49S GKOMKTRV. APl'ESinX
Hippasus {470 B.C.) discovered the dodecaliodrnn. but,
he was drowned by the other Pythagoreans for boasting
of the discovery.
Eudoxus (380 E. C.) showed that the volume of a pyra-
mid is equivalent to one-third the product of its base by
its altitude.
E. F. August (Germany, 1849) introduced the prisma-
toid formula into geometry and showed its importance.
855. The Three Round Bodies. Eudoxus showed that
the volume of a cone is equivalent to one-third the area of
its base by its altitude.
Archimedes discovered the fonuulas for the surface and
volume of the sphere.
Menelaus (100 A. J).) treated of tlie properties of
spherical triangles.
Gerard (Holland, 1620) invented polar triangles and
found the formulas for the area of a spherical triangle and
of a spherical polygon.
856. Noa-Euclideaa Geometry. The idea that a space
might exist having different properties from those which
•we regard as belonging to the space in which we live, has
occurred to different thiuiers at different times, but
Lobatchewsky (Russia, 1793-1856) was the first to make
systematic use of this principle. He found that if, instead
of taking Geom, Ax. 2 as true, we suppose that through a
given point in a plane several straight lines may be drawn
parallel to a given line, the result is not a series of absur-
dities or a general reductio ad absurdum; but, on the con-
trary, a consistent series of theorems is obtained giving
the properties of u space.
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ITT. REVIEW EXERCISES
EXERCISES. CROUP se
REVIEW EXERCISES IN PLAXE GEOMETEY
Ex. 1. It the bisectors of two adjacent angles are parpendioulaj
to each otber, the angles ai'e supplementary.
Ex. 2. If a diiigonal of a quadrilateral bisects two o£ its angles,
the diagonal bisects the quadrilateral.
Ex. 3. Through a given point draw a secant at a given distance
■om the center of a given circle.
Ex. 4. The bisector of one angle of a triangle and of an exterior
ngle at another vertex form an angle which is equal to one-half thtj
lird angle of the triangle.
Ex. 5. The side ol a square is 18 in. Find the ulrcumference of
the inscribed and eiroumseribed circles.
Ex. 6. The quftdrilateral jlDiJC iK inscriHeii in a circle. The diag-
o.iiils Af! and iJCintersect in the point F. Arc ^J> = 112°, Rrc AC =
lAFC=^ li". Find all the other angles of the figure.
Ex. 7. Find the locus of the center of a circle which touehee two
given equal circles.
Ex. 8. Find the area of a triangle whose sides are 1 m., 17 dm.,,
210 em.
Ex. 9. The line joining Ihe midpoints of two radii is porpendiculaf
to the line bisecting their angle.
Ex. 10. If a quadrilateral be inscribed in a circle and its diag-
onals drawn, bow many pairs of Bimilar triangles are focnied t
Ex. 11. Prove that the sum of the exterior angles of a polygon
(Art. 172) equals four right angles, by the use of a ligure formed by
drawing lines from a point within a polygon to Uie vortices of the
polygon.
(430)
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GE05IETEY.
Ex. 12. In a circlo whose radius is 12 em., find the langth of the
lugent drawn from a point at a distance 240 mm. from the center,
Ex. 13. If two sides of a regular pentagon he prodiieed, End the
Ex. 14. In the parallelogrEm ABCD, points are taken on the
diasoaals such that AP=BQ=CE==DS. Show that PQBS is a
parailelogram.
Ex. IB. A chord in. long is at the distance 4 in. from the eentcv
of a circle. Find the distance from the center of a aiiord 8 in. long.
Ex. 16. If B is a point in the eireumferenee o£ a eicele whosa
center is 0, PA a tangent at any point P, mooting OB producBd at .-),
and PD perpendicular to OH, then. PB bisects the angle AFD.
Ex. 17. Construet a parallelogram, given a side, an angle, and a
diagonal.
Ex. 18. Find in inches the sides of an isosceles right triangle
whose area is 1 sq. yd.
Ex. 20. 3f two lines intersect so that the product of the segments
of one line equals the product of the segments of the other, a cir-
cumference may he passed through the extremities of the two lines.
Ex. 21. Find the locus of the TerticBB of all triangles on a given
base and having a given area.
Ex. 22. On the figure p. 206, prove that^'*+4p = ZB^H-f(^^-
reetangle is 108 and the base is throe times
Ex. 24. If, on the sides AG and BC o£ the triangle ABC, the
squares, AD and BF, are constructed, AF and DB are equal.
Ex. 25. If the angle included between a tangent and a secant is
half a right angle, and the tangent equals the radius, the secant
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REVIEW EXERCIHES IN PLANE GEOMETRY 501
Ex. 26. The sum of the areas of two circles is 20 sq. yds., and the
difference of their areas is 15 sq. yds. Find their radii.
Ex. 27, Construct an isosceles trapeaold, given the bases and s,
Ex. 28. Show thfit, i£ the alternate sides of a regular pentagon
he produced to meet, the points of intersection formed are the -vertices
of another regular pentagon.
Ex. 29. If a poat 2 ft, fi in. high caats a shadow 1 ft, 9 in. long,
liow tall is a tree which, at the same time, casts a shadow H6 ft. long I"
Ex, 30. If two intersecting chords make equal angles with the
diameter through their point of intersectiju, the chords are equal.
Ex. 31. From a gireii point dra
einal segment Is half the secant.
rele whioh touehus
Ex, 33. If one diagonal of a quadrilateral bisects the other
diagonai, the firat diagonal divides the quadrilateral into two equi-
Ex. 34. In a given square inscribe a square having a given side.
Ex. 35. A fleld in the shape of an equilateral triangle contains
one acre. How toaiij feet does one side contain f
Ex. 36. IE perpendiculars are drawn to a given line from the ver-
tices of a parallelogram, the sum of the perpendiculars from two
opposite vevtiees equals the sum of the other two perpendiculars.
Ex, 37. Any two altitudes of a triangle are reolproeally propor-
tional to tbe bases on which they stand.
Ex. 38. Construct a triangle equivalent to a given tj-iangle and
having two given aides.
Ex. 39. The apothem of a regular hexagon is 20, Find the area
ot the inscribed and circumseribBd circles.
Ex. 40, M is the midpoint of the hypotenase A£ of e, right tri-
angle ABC. Prove 8 ilC~ = jJs"+llC~+Tc''-,
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50'-i C.KOMF/l'llY. ArrHNlHX
Ex. 41. Traiififonn a civon tvianjjlo into an e
angle eontaiuijig a given acute angle.
Ex. 42. The area of a square inscribed in a
aiea of the square inscribed ia the oirole as 2 : 'i.
Ex. 43. If, on a diameter of the oirele 0, OA-
aliel to BD, the chord CD is perpendicular to JC,
Bi. 4-5. State and prova the converse of Prop. XXI, Bonk III.
Ejt. 46. If, in a giren trapezoid, one base is three times the otlifc
base, the segments of each diagonal are as 1 : .t.
Ex. 47. If two Bides of a triangle are 6 and 12 and the anglt-
included by them is liil°, find the length of tlie other side. Also lind
this when the iuciMdod angle ia 4."!° ; also, when ll!0^.
Ex. 48. How many aides has a polygon in whieh the sum of the
interior angles exceeds the siita of the exterior angles by 540"?
Ex. 49. If the four sides of a quadrilateral are the diameter of a
oirele, the tvo tangents at its extremities, and a tangent at any other
point, the area of tlie quadrilateral equals one-half the pcoduet of the
diameter by the side opposite it in tlie quadrila-teral.
Ex, 50. An equilateral triangle and a regular he.iragon have the
Bsrae perimeter; End the ratio of their areas.
Ex. 51. To a circle whose radius is 30 e
from a point 21 dm. from the c
Ex. 52. If two opposite sides ot a quadrilateral are equal, and
the angles which they make with a third side are equal, the quad-
rilateral is a trapezoid.
Ex. 53. If two circles are tangent ostornally and two parallel
fliametera are drawn, one in each circle, a pair of opposite extremities
of the two diameters aud the point of contact are eolllnear.
Ex. 54. If, iu the triangle ABC. the line AD is perpendicular to
BD. the bisector of the angle II, a line through V parallel to BC
bisects AC.
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EKVIFAV EXERCISES IN PLANE GEOMXTRY i)0-i
Ex. 55. BLaocl a given triangle by a line parallel to a given liae.
Ei. 56. It two parallelograms have au angle of one e-^uul to the
Eupplement ot an angla of the other, their areas ai'o to each other aa
the produetB of the sidoa inoluding the angles.
Ex. 57, The sum of the medians of a triangle is less than the
perimeter, and greater than half the perimeter.
EiC. 58. If PARIS is a secant to a <>,irele through the center O, FT
a tangent, and Tit perpendicular to Pli, then I'.l ; l'lt = I'0 : PIS.
Ex. 59. Two concentric circles have radii of 17 and 15. Piud the
length of the ehord ot the larger which la taugtrnt to the smaller.
Ex.60. Onthe egure, p, 244,
(o) Find two pairs of similar triangles;
(6) Find two dotted lines whieli are pei'pendiculac to each other;
(e) Diaeover a theorem oout-erning points, not connected by lines
on the figure, which are eollinear;
{(J) Discover a theorem concerning sqnares on given lines.
Ex. 61. One of the legs, AC, of an isosceles triangle is produced
through the vertex, C, to the point F, and F is joined with I), the mid-
point of the base AB. I)!'' intersects IlC in £. Prove that Cf ia
greater than CE,
Es. 62. The line of centers of two circles intersects their common
external tangent at P. PAliCI) is a secant intersecting one of the
two circles at A and U and the other at C and D. Prove PA X PD =
PEXPC.
Ex. 63. Trisect a given parallelogram by lines drawn throut'h a
given vertex,
Ex. 64. Find the area of a triangle the sides of which are tho
chord of an arc of 120° in a circle whose radius is 1 ; the chord of an
arc of UO" in a circle whose radius is 2; and the chord of an arc of
60° in a circle whose radius ia 3.
Ex. 66. Two circles intersect at P and V. The chord CQ is t,
gent to the circle QPIl at Q. APB is auy ehord through /'. Pn
that AC is parallel to Bq,
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Ex. 67. la tho trianglo ABC, from 7», tlio midpoint of Jli; J
and DF are firiiwn, biseetlng tbe angles ATJIl and ADC, and maeli
AB at E and AC nt F. I'rove £F || ISC.
Es. 68. Produce the side KC of the triangle ABC to a point />,
that PBXfC=I^j'.
Ex. 69. lu a giveu circle inscribe & reetan^le similar to a f;i\
rectangle.
Ei. 70. In a given sonii
.cin'lo i
ns.:illji) a rtcliinjjiii similiir to ?i
given rectcngie.'
Ex. 71. The area of an i
soSL'oles
trapcKoid is 140 aq. ft., one bas«
la ^6 ft., and the legs make <
in angle
or 4:,- Willi tlic other liaae. Find
the other base.
Ex. 72. Cut off one-thli
rd tho a
rea of a given trinngie by i, iin^
perpendicular to one side.
Ex. 73. Fiod tbe sidea
of a tri^
ingl^ wLo-^o area is 1 sq. ft., if
the sides ave in the ratio 2 :
: :i : 4.
Ex. 74. Divide a given line into two parts such that the aum of
the squares of the two parts shall be a minimum.
Ex, 75. If, from any point in tbe base of a triangle, lines are
drawn parallel to tbe sides, find the loeus of the center of tbe paral-
lelogram ao formed.
Ea. 76. Three siiios of 'a quadrilateral are 845, 613, filO. and the
fourth side is pei'pendieular to the siijes H45 and HIO. Find the area.
Ex. 77. It BP bisects the angle ABC, and BP
bisects the angle CHA, prove that angle F=i sum
of angles A and C.
Ex. 78. Two circles intersect at F and Q.
Through a poiut A m one oireumferenoe lines Al'C
and AQD are drawn, meeting the other in C and 1). Pr
gent at A parallel ta CD.
Ex. 79. In a given triangle, draw a line parallel to tbe base and
terminated by tho aides so that it shall bu a mean jn'oportional be-
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REVIEW EXEIiCISES IN PLANE GEOMETRy 505
Ex 80. Find the angle iasoribed in a asmioirole the Bum of whose
rides is a maximum,
Ex. 81. The bases of a trapezoid are IGO and 120, and the alti-
tude liO. Find the dimensions of two equiTalent trapeioids into
which the given trapezoid is divided by a line parallel to the base.
Ex. 82. If the diameter of a given circle be divided into any two
segraente, and a seraioireumfereQce be described on the two segments
on opposite aides of the diameter, tlie ai'ea of tlie civcle will bo di-
vided by the semieii'cumferencea thus drawn into two pai'ts having
tlie same ratio as the segments of the diameter.
Ex. 33. On a given straight line, AB, two segments of oiroiea are
drawn, ^PJSnml AQH, Tlieanglea QAI' tmiQBF are bisected by Unas
meeting in E, Prove that the auglo £ is a constant, wherever P aiiJ
Q may be on their arcs.
Et. 84. On the side AB of the triangle ABC, aa diameter, a cir-
cle is described. AT is a diameter parallel to BC. Show that EH
bisects the angle ABC.
Ex, 85. Construct a trapezoid, given the bases, one diagonal, and
an anfjlo included by the diagonals.
Ex. 86. It, through any point in the common chord of two inter-
aectinf; circles, two obonls be drawn, one in each circle, through the
four extremitiee of the two chords a ciruumference may bo passed.
Ex. 87. Prom a given point as center dsserihe a circle cutting a
given straight line in two points, so that the produot of the distances
of the points from a given point in the line may equal the square of a
given line segment.
Ex. 88. AB is any chord in a given circle, P any point on the
cireumference, P\[ is perpendicular to AB and is produced to meet
Iha circle at Q; AN is drawn perpendieular to the tangent at P.
Prove the triangles A'AM and PAQ similar.
Ex. 89. If two circles ABCD and EBCF intersect in B and a and
have common exterior tangents AE and D-F aut by BC produced at <j
audi/, thi^u VH'- = Ba--\-AE-.
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aUh GKOMKTKY. Al'fENiilX
EXERCISES. CROUP BT
RKVIi;\V EXERC'lfiES IN SOl.lD liEOMETRY
Ex. 1. A sepment of a straiRht line oblique to a piane is greatei-
than its projeclioti on Ihe plane.
Ex. 2. Two tetvahedroiiri are aimilar if a dihedral angle of oim
equals a diliedral angle of the other, and the faeea forming tboso
difieilral angles are similar each to each.
Ex. 3. A plane and a straight line, both of wliic'h are parallel tc.
the same line, are parallel to each other.
Ex. ,4. It tlie diagonal of one taue of a cube is 10 iurlits, find Iht,
volume of the cube.
Ex. 5. Construct a spherical triangle o
■n a ijiv
lea of the sides of the triangle.
Ex. 6. Given AJ! 1 MX,
Jfand KF I. MR;
prove EF X I'M.
Ex. 7, The diagonals of a reetangula
r paral-
Ex. 8. What portion of the surface of a sphere
is a triangle eseli of whose angles is 140"?
Ex. 9. Through a given point pass a plane parallel to two given
straight lines.
Ex. 10. Show that the lateral area of a cylinder of revolution is
equivalent to a circle whose radius is a mean proportional between
the altitude of the cylinder and the diameter ol its base.
Ex. 11. The volumes of polyhedrons circumscribed about equal
spheres are to each other as the surfaces of the polyhedrons.
Ex. 12. Find Sand T of a regular square pyramid sn edge of
whose base is 14 dm., and whose lateral edge is 250 cm.
Ex. 13, It two lines are parallel and a plane be passed through each
hese pianes is parallel to the given lines.
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EEVIEW EXERCISES IN SOLID GEOMETRY
Ex. 14. Givoii Pll X plane AD,
Z rEU= Z FFH ;
prove lI'El'=Z.rFE.
El. 15. If a plniie be passed
through the midpoiiila of three
edges of a pnrallelopiped wliiirh
meet at a vertex, the pyramid thna formed 13 what purt of the
parallelopiped f
a of its dist.anp«s
Es. 17. Given the points A, Jl, C, I> in a plane and Pa point
outside the plane, J/1 perpendieuldr \o the phme PBD, and AC p.T-
pendiKular to tlie plane i'VIi; prove lliiit I'D Is perpeudieuhir tu
the plane Al'.CIi.
Ex. 18. In n spherb whoso radius is ">, find the aifii of a /nnti tiju
radii of whose upper and lon-er hases are -T and 4.
Ex. 19. Two cylinders of revolution havK equal lateral aroaK,
Show that their volumes are as It : It'.
Ex. 20. The midpoints o£ two opposite sides ot a quadrilateral in
space, and the midpoints of its diagonals, are tlie vertioes ot pi
parallelogram.
Ex. 21. How many feet oE two-ineh planlt are necessary to eon-
Btrucf. a box twice as wide as deep and twice as long aa wide (ou the
inside}, and to contain 216 on. ft.!
Ex. 22. If two spheres with radii K and r are concentrie, find the
nreu of the section of the larjjer sphei'e made by a plane tangent to
the smaller sphere.
Ex. 23. In the frustum of a regular square pyramid, the (!.!;.■■■»
of the bases are denoted by b\ and hi and the altitude by H ; pvusu
that X=iv'(6i— i-il'-h^fl^
Ex. 24. If the opposite slJ
V .>|..posite angles are equal.
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508 (jEOiii'iTKY . APPi;>; dix
Ex. 25. Obtiiin the simplest formiita for the lateral suvfaee of a
truDeated tvianfrular right prnm, each edge of whosf base is «, and
whose lateral edges are 21, 9, and r.
Ex. 26. The area of a zone of one ba^o is a mean proportional
between the reiuainiiig surface of the sphere and its eiitiio surface.
Find the altitude of the zone.
Ex. 27. The lateral edges of two similar frusta are as 1 ; <i. How
do tiieir areas compare ? Tlieir volumes f
Ex. 28. Construct a spherical surface with a given radius, r, which
shall be tangent to a given plane, and to a given sphere, and also pasa
through a given point.
Ex, 29. The volume of a right circular cylinder equals the area
of the generatiuf- rectangle multiplied by the ciroumferenee generated
by the point of intersection of its diagonals.
Ex. 30. On a sphere whose radius Is Si inches, find the area of a
zone generated by a pair of compasses whose points are 5 inches
Ex. 31. The perpendicular to a given plane from the point where
tlie altitudes of a regular tetrahedron intersect equals one-fourth tho
sum of the perpendiculars from the veitieee of the tetrahedron to the
same plane.
Ex, 32. Two trihedral angles are equal or syminetrieal if their
corresponding dihedral angles are equal.
Ex. 33. On a sphere whose radius is a, a zone has equal banes
and the sum of the bases equals the area of the zone. Fldd the alti-
tude of the zone.
edges of a tetrahedron
Ex. 35. Find the loeus of all points in space which have their
istances from two given parallel lines in a given ratio.
Ex, 36. It a, b, c are the sides of a spherical triangle, a', 6', c*
iie sides of its polar triangle, and a>b>c, then a'<l/<e'.
Ex. 37. A cone of revolution has a lateral area of 4 sq, yd. and
n altitude of 2 ft. How much of tho altitude must be cut oft by a
ilane parallel to the base, in order to leave a frustum whose lateral
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EEVIEW EXERCISES IN SOLID GEOMETKY
Ex. 38. The total a
isoribed sphere as 9 : '
Ex. 39. CoQstrucl a sphere of give
be thngent to three given, spheres.
eu. in. and its altitude ia 20 in. Show how to find the radii of the baaes.
Ex. 41. On each base of a cylinder of revolution a cone is placed,
with its vertex at the center of the opposite base. Find the radius of
the circle of intersection of the two oonieai surfaces,
Ex. 42. The volume of a frustum of a cone of revolution equals
the sum of q cylinder Bud a eone of the same altitude as the frustum,
and with radii which are respectiveiy the half sum and the half differ-
ence of the radii of the frustum.
Ex. 43. A square whose aide is a revolves about a line through
(mo of its vertices and parallel to a diagonal, as axis ; find the surface
and volume generated.
Ex. 44. If a cone of revolution, roll on another fixed cone of revo-
lution so that their vertices coincide, find the kind of surface gen-
erated by the axis of the roiling eone.
Ex. 45. An equilateral triangle whose side is a revolvea about au
altitude as au axis; find the surface and volume generated by the
inscribed circle, and also by the eireumseribed circle,
Ex. 46. Find the locus of the center of a sphere which la tan-
gent to three given planes.
Ex. 47. If an equilateral triangle whose side is a be rotated about
a line through one vertex and parallel to the opposite side, as an axis,
find the surface and volume generated.
Ex. 48. What other formulas of solid geometry may be regarded
as special eases of the formula for the volume of a prismatoid ?
Ex. 49. Through a given point pass a plme which shall bisect the
volume of a given tetrahedron.
Ex. BO. In an equilateral <
leuts are porpetidicuhir at tl
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PRACTICAL APPLICATlOnS OF PLANE GEOMETRY
EXERCISES. GROUP S8
(Book I)
1 Take 1 piece of pipir li urns i '^IiiirI
o iaim I !i[;lit aiit,k \\ hjt gtomi I
( liGO inil fol 1 thp paper
I |iriiKi]ik en definition
Wh^t RComHru
.i.nuplK
hu
^ou
nU)cout'iii!eanp:l
halt the anfrle, j:
thp dijgruni (tL
..f thP c
llPt«Cl
note the
■upp
the
iteis
out
bye
l^e It IS csMutull^ the mrfhod wfd
pe ind hcn(u m corn ctijig inslni
itiaiuhL pcli,!, test the j,ipuiii\ of the
I iiippJifLi^ -jqiiri ljy u meth'xl
he tliiigr nil Ilo« thtn would
ipLUi in of tin 111-kU uirIp of
3 InE\ iproiethit lh( PI
square if there he auj cqii il« o
side lines of the square as sho«n ii
and show that e + r = j;)
T-his prmeiple la irapoitant btea
in correcting the a\is of i teloioc
menta of which thp t Icicopt h a part ;
HUrieving and astronomical instruments
4 By use of a ciipenters square am
sltaight edge lay off i strits of pnnllel lines \\hat
property of pii lilel line^ ba^e joii used'
5 Tell how to (,onBtru.ct a carpcntoi i miter box
6 Ihe (ii^run hhows a di miiiK instr imuit edled
thepirill 1 lulir, lh( dotud outlm
of thi. in-.tiuni(nt in another jiosilion th pirt RS
leimining faxiil Thp dist mci i'Q = hS, PR = Q'^
HeiiLL show thit PQ ii patalltl to R!>
In like manner show thU PQ i» imralld to RS
Hi nee Awv. PQ i^ panllLl to P Q
If \ line b( ilriwn p( rp< nditular to PQ and
anothrt p(ip<ndicuUr to P Q the pctpcndiiul ir ll u~ d i
pmllci to lach i.thir { \it 111 lin,-, i Liptu hi. In to i , ill Uii i.
art, parjlkl)
510
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PHACTECAL APPLIPATIOXS f)!!
The above are ctbcs of linked motion, a kind it mMhanism o[ mdo
importance Observe fonnatance the system of hnks which oonntLt
the driving wheels ot j loLomotivi nith the [iiaton in the tyhnder and
also the jointed rods conntctiug thi walking bi im ot j, atearaboa! nith
the engine Look up ilso, m the Cmliuv DictionavVi for instuncL
the words hnkagt, icii, and jiaialkl motion
7. The distdnec- bpifliiii
paflaable bainer (h» b
a pond) may be Idi i
no instrument for in i
Ut A and B Ix I hi >
iMit Bliitioii ( a,nd meiaiirL
to F, making ( I = l<
DC = liC Mtasiirt tJl
1! AC = Cr -220 ft,; liC -
long is AB?
8. In the trusses of steel bridges, why are the beams nnd rods
arrai^ed so as to form a network of triangles as far as possible, and not
ot quadrilaterala, or pentagons, for instance? (See Ex. 3, p. 70; also
Art. 101.)
How 13 this principle also made useotin forming the frame of a wooden
hiiX'^Q oiboxcir or tostren^thni i « eak f i ime < r f ener of an\ kind'
9 Draw a map for the lollowmg sar\(j Hot's to the seilo ot
400 ft to lh( inch In Hjitir!
off the an^k'i dia" i. dott(.d
lorth and south line through
a<!i stitioti ml then use a
l^^o iCHs-ii>li ph(e,
sepaiated bi in
m tlioi
1 luml
L l( iiadHI I'lodi
I'rodULG m U. I> 1
;;,:■;. .^^.
' " / \
,1 1,11 u.
Phul lis = 1)1
liC = Cl> -liiOtL;
ami IJF - 21(1 fl., 1
Stations
|„,
iiing=i
Di,l mceh
1
JN
"0 i
m> fl
1>
' \
J)i) fl
1 s
r"}
JiOfi
' D
1
■^
.o°\\
bOOlt
Kttp your dialing £or a
iltr iiw
10 Obtain oi nnkeupawLt
ot Bur\e\ nolcf, snnd ir lo those in E\ md niakt a lii using tor them
U IitD( and/Tfp ol2l be tno naih iRiiMndiiiilar to the ]>Iiti(
fthtpipti Lit smiUmnrorslji' itluhi-d to those » Ulh at land/)
Ul Z( - W I,,turi\ of hght pi«i Ihiough y 1o 1 bircflul I
I 1 /i inclthimoto/' ProvctSiil/ l/'B is i right ingle
[->LC The Ian of reflicted light is that the ai^le of inLideni e e(]n i\-
thi angit of reflection or in the figure x = x and u = y ili(.u Jit
thi triinuln ABC j; + ,j + i^" - ISO'' or i + y = n-i° \l out thi
p, iitB 1 ani b it i-'t +21/ + 1 = 100° u + d = tJO ttc ]
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(";eometry
Bht c
gf
t s Kbt Rl 1
II h 1 113
t gl thth
ray coming from Q through /' to A.
IS, The velocity of light is dolermined by the
' " in till! following manner:
J_ pI.TJie of the paper ami
livot; let OP, bel,-l,fi;.
it and reflected Ihroujih
ills distant, whence it i>
retuni of fhe ray to 0,
a rotating
Let A,Bi be a mirror
rotating about as u ]
Let to be a ray of light striking the miritir :
M to a small Rtafionnry mirror some m
ref)i:ttcd back through .1/ to 0. On the
A,B, will have rotated through a
small angle, o, to the position AiB-;
hence the ray will be reflected in the
direction OR, Let the pupi! show that
Za = I^Z LOR. Since Z. WR may
readily be measured, ^ a is known, and
if the rate at which the mirror is rotat-
ing is known, the time occupied by the
ray in traveling from to the stationary
[Suo, ^PiOPj = Z /l.O^! (Art. 132)
Z_P,OL = Z. PvOM = cc ( / incidence = Zreflcfliuti)
.■- Z P^OL = x~a
.■. ZiOfi =» + a- (x-a) =2al
If Z LOR = 2° 19', OM = 3 mi., and the mirror AB rotates 100
times per second, determine the velocityof light per second.
13. To prove that the image of a point in a plane mirror is on a
perpendicular from the point to the mirror and as far behind t!io
mirror as the object is in front of it, let
MM' be the mirror and P the point, and
./> PAE and PA'E' two reflected rays. Show
and back Js determined.
Z PAM = AEAW = £ P'AM.
:. Z /M.U' = Z P'AM'.
. PAA' = 4 P'AA'.
.-. PA = P'.i, etc.
14. From the above show the direction in which a billiard ball must
be sent in order to strike a certain point on the table afler striking oat
side uf the table.
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p
h
f
\
7 /
b
/
-'
PRAmCAL VPPLI'^ATION'* ■51'i
lors 0.1/ anil 01/ bp peiprnd :hr to pt h (thpr
makp i pon'if ruction to show
nhere the point P mtil appear
to be to an e>e at E aftir tbe
light from P hia hppn iifltctpd
from both mirrors
hoc Pio\eSP'' = A 1 + 4P
= B 1 + IP ]
16 ■\\hat woull liL thL ippU-
i ition of F^ 15 to a ball -truck
jn a biUiaid tabU
17 The path of a ni} of light
before and iiftcr enlcnnji; a Rlaia
prism IS gnen b\ ihc Imes AB
and CD The entire angle bj «hich a
raj of light IS defieetpd on passing thro igh
the prism is denoted by x Prove that
J- = , + r - P (Zi T'Bn and PC^aie 1 1 ZO
ta. By use of d quadrilateral BPCy,
y = 180° - P. Then use the quadrilateral
all of whose sides are dotted lines,]
EXERCISES. GROUP i
2, By use of the c.irpenfer's sqiiaro find the
center of a given circle.
3. Show how a. pattern maker by the use of a
carpenter's square can determine whether the
ciivity made in the edge of a boatd or piece of
metal is a Beniicirclo,
4. Biboct a (civcn angle by the use of a carjientpr's sqiiaic.
fi. By use of squared paper divide a line U inches long into five
equal parts. Info 7 equai parts. Into 3 equal parts. Can you make
this division on paper ruled in only one direction?
6. Make tip ami work a similar exatjiple tor yoursolf.
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(GEOMETRY
long and let be ifK midpoitiL Murk off
four equal parts at D, F, li. With O
through C, Z), P, H. A. From draw
radii (one of which is OA) dividing the
angular space about into eight equal
pxrts Beginning at dran tht
smooth curve ( VHi-tb through thp
joints where the ladii drawn inter
ect the circumferpin.es ui, indieatcil
m the diagrim Ihc lesult will be
the outline of i special foim of wheel
(ailed a C'jm much UM'd in machine
t mo-
th g aid
] ea p t b Pe
g 1 a;
1 l g hoes A ;
1 t t
back and forth motion. TI i ff f 1 g h bet
on the diagi'am ia termed tl ti f th
8. Make a lirawing of } h th thi h
the longest radius IJ in.
9. Stretch a 100 ft. tape or part of it so ai
10. To extend a straight line AB beyond an obslatic, ;is a building,
we may proceed as follows:
At B measure off an angle C ,.
ABC = 60°. Produce CB to D, /\ /\ \
etc. Let. the pupil complete the a — '— ^^ /^ — ^ — —K
construction and prove lliat his \ /
method is correct. V
11. Let the pupil solve the
problem of Ex. 10 by the construction of right angles instead of angles
of 60°.
12. To find the distance between two points A and B, one of which
A l"1 ^■^^ '^ inaccessible, we may proceed as fol-
p _ u--_j,B idYjg: Extend BA to C. Measure a con-
Q/^''%\ venient line AD and extend AD to E,
^^^-^y f| making DB = ^D. From £ run a line 1|
.F^ E \\. ■^'' ^""^ meeting BD extended at F.
^ Measure £f, Prove that £/■ = A£.
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PRArXICAL APPLICATIONS
515
if'
13 Show how to solve the problpni of F\ 12 iij' c-onsfriieting a
hne at right angles to xnc Iher hne instead of one parallel to inotlier,
14 If two Eirfetfi meet as in the diagram
show hoB, a cur\o of gi\en radius r may be
made to tike the ylaee of the angle 4 and be •
tangent with the curb of the two btveets
16 The curve in a railroad track is usually
ao arc of a circle tangent to each straight track
T which it joinn If two straight fra^-l « 4B and
r>. '"D oto joined bj a circuUr curve tangent to
I ofh Dt them and P la the point where IS and
( D \o !J mft it e\tKnded piove ZTI D = 2
Z. Tii
16. One na> of Ujmg off a railroad curve
n track is as follows; Let ABP be the ^
given straight track. At B construct a small angle PBC
(whose size depends on the degree of curvature which
the curve is to havet. On it mark off BC = 100 ft. y
Construct Z CBD = Z P^C. Take C as a center and 7
100 ft. as a radius, and describe an arc cutting BD /
at D. Construct E from D in like manner. Prove /
that, B, C, D, B all lie on
/ the arc of a circumference tangent
at B.
[Sua. Pass a circumference through B,
C, D. Prove that B lies on this circumfer-
^e and that ABP is tangent to it.]
17. Let AB and CD be two straight rail-
O road tracks connected by an arc AF of a
circle whose center is 0, and an aic FC
whose center is 0', the arcs having a com-
mon tangent at F. Prove that the angle of
n (z) of the two straight tracks, if produced, equals the sum
of the central angles of the two tracks.
[Sue. X ~y + z. Use Ex. 15-1
A curve like AFC composed of two or more arcs of different radii is
called a compound curoe. Can you suggest why a compound curve
should lie used in connecting railroad trajjks?
18. If Hip two arcs which compose a compound curve lie on opposite
sides of .their common tau^i;iil, the comjiounii curve ,ia called a reoersa
) AB
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GEOMETRY
Thus on the diagram, BCD
nnecting the
itraight roads AB and DE.
Discover the relation between
\% Whiiij is ii railroad Irogl' If a ciirvod track c
track, show that the angle of the frog (i) equals the
central angle of the curved track (o).
20. If two tracks which curve in the same direction
crosB each other, find the relation between the anglo .
o[ the frog and the central angk-s of the two curved tracks.
21. Find the same when the two curved tracks curve in opposite
dii'ontions.
22. Show how to locate a gas-generating plant (P) along a given
j^ straight road AB so that the length of pipe
/ connecting it with two towns (C and D)
"■■v^ / shall be a minimum.
"'-./ , (Use Ex, 24, p. 176.)
23, Discover and state ;;
Ex. 24, p. 17U, similar to one given in Ex. 22.
24. If a treasure has been buried 100 ft. from a cert;
distant from a given straight road and a path parallel t(
how the treasure may be found.
26. Make up and work a similar example for yourself using the prin-
ciple of E)i. 2, p. 167, Also using that of one of Exs. 3-S, p. 168.
26. Prove that the latitude of a place ^n the
earth's surface equals the elevation of the pole
(that is, on the diagram, prove Z. OE-4 = Z.PAO).
„ Z_ 27. Given the sun's declina-
tion (i.e. distance north or south
of the celestial equator), show
how to determine the latitude
^ ^ ' of a place by measuring the zenith distance of the
sua- Also by measuring the altitude of tb.e sun above the boiison.
application of
o the road, show
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PRACTICAL APPLICATIONS
517
3 and 27 i
28, How was Peary aided by the principles of Eks.
determining whether he had arrived at the North Pole?
29. If on April 6 (the day of the year on which Peary was at the
North Pole) the sun was 6" 7' north of the celestial equator, how high
above the horizon should the sun have been as observed by Peary?
At what hour of the day was this?
80. The sextant is used almost daily by every navigator in deter-
mining the altitude of the sun above the horizon, and hence the latitude
of the ship. The construction of the sextant ia ^
based on the principle that if a ray of l^t be
reflected from two mirrors' in succession, the
change in the direction of the ray of light
equala twice the angle made by the mirrors.
Prove this law.
|Su(j. Let M and N be the mirrors and the
reflected ray be SMNO. It is required to
prove that Z.y = 2 ^x. It is to be noted
that Z.FNM, being an exterior angle of triangle
NS'M, equals x + i. Hence angle MNO equals \ /
180° -2 (e -l-t). Hence in triangle MNO, \^/
y + ISO" - 2 (a; + t) + 2 1 = 180°, etc.] \^.
In the sextant, angle y is the elevation of thi"
sun above the horizon, and x can be read on the rim of the u
1 position V
mirror N i» fij.ed i
M rotates
It maj be observed thit the prin
fjple of this example is the ^an < la
that pro\ed m E\ 12 p 512
31. The diagiim show? a some-
what simple aubstitutp tor the e\
tant called the angle meter. ih the
center of the arc BC and MM' is a
fixed mirror. The instrument is
held BO that a ray SO from the sun
when reflected from the mirror passes
into the eye (A) in a horizontal direc-
tion. Show that the elevation of tliE>
The rim BC is graduated so (hat
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ol8
GEOMETRY
The angle-meter may be used to
angles.
What is the advantage in using a mirror on
sort? That is, why is not the positioD of the
across a graduated arc?
32. Draw an easement &
liorizontul as well a;
instrument of this
n observed directly
i composing the
e (.AB) tangent to the rake cornice Bc
and passing through a required point A. (The
e construction is used in layiii^ out the ease-
ments of stair rails, etc.)
(Use Ex. 7, p. 174.)
33. A s^mental arch is a coraijound curve
composed of the arcs ot thriie circUs. The method
nof constructing a segmental arch is as follows;
Let AB be the span and CD the altitude of the re-
quired arcli. Complete the rectangle GADC.
Draw the diagonaJ AC. Bisect the angles
GAC and GCA. Let the bisectors meet at E.
Draw EH perpendicular to AC, meetii^ AB at
JV <md CD produced at U. Make DK equal to '^'
DN. Then show that N is the center and A'.'l
the radius, H the center and HE the radius,
and K the center and KB the radius for the a
arch. (See Hanstein's Constructive Drawing.)
[ScG. At E draw a hne perpendicular to EH. Then Z. -VB-1 -
Z. NAE. (Complements of equal angles are equal) etc.]
34. What is called a Persian arch may be constructed as follows;
Let AB be the span and CD the altitude of the required arch. Draw
p n P *he isosceles triangles ADB. Divide AD into three
equal parts at H and G. Construct HK ± AG
at H, and meeting AB produced at K. Produce
KG to meet Ef which has been drawn through D]|
AB. With K as & center and KA as a radius, and
. B as a center and EG as a radius, describe arcs
meeting at G. Prove that these area have a com-
mon tangent at 0, and therefore form a compound
5urve. (See Hanstein's Constructive Drawing.)
3B. Construct a Persian arch in which the arc DG = arc NG.
[Soa. Bisect line AD instead of trisecting it.]
36, Construct Persian arches in whicli llio chord:; NG and DG have
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PRACTICAL APPLICATIONS 519
various ratios, and decide wliich of these arches you think is the most
beautiful.
37. Construct aegmental arches of various shapes and decide which
of these you think is the most beautiful.
38. Construct a trefoil, given the radius of one of its circles.
[SoG, Construct an equilateral triangle, one of whose
sides equals twice the given radius. Take each vertex
of Ihe triangle as a center, and half of one of the sides
iw a radiua and describe arcs.l f ^
39. Inscribe a trefoil in a given equilateral triangle, V_yV^_y
[S0G. Bisect, the angles of the tiiu,ngle, etc.]
40. Discover the method of eoiislrucdou of each of tlu' figuvpr! or
diagrams on the next page, and then rcpi-oduce each <•( tiicm in a
drawing.
Squared paper may be used to advantage as an aid in making some
of these constructions. See Fig. 10, p, 520.
41. Construct a diagram siiailar to Fig. 10, but making a rectangle
instead of e, square the basis of the drawing.
42. .\lso one making the rhomboid the basis.
43. Construct a trefoil and develop into !Ui ornamental design by
placing small circles and ares within its parts and larger circles outside.
44. By use of squared paper invent designs similar to Fig, 10 on
p. ,')20, but formed by two intertwining lines,
46, Collect a number of pictures of ornamental designs, tracery,
scroll work, etc., such as arc vised in architecture, wall paper, and
similar patterns, whose construction depends on the principles of Book
II. Show how these designs are constructed, and reproduce them in
drawings.
EXERCISES. GROUP 90
(Book III)
1. To tind the distance between two accessible
objcpt.-i ;1 and B separaliid by a bari'iet', take a
convenient point C, measure AC and BC. Protluce
EC to I) and AC to F so that DC = some frac-
tion, as i, of CB, and FC = the same traction of
FD. How, then, is All com-
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rmre
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PUACTiCAL APPLICATIONS
521
Let BC = 300 yd., DC = 60 yd,
FD = 52 yd., find .IB.
Why is this method of finding AB
of Ex. 7, p. 5U.
2. What similar method doea the
gram surest for finding AB?
3. In the di.igriim of Ex. 2,
between A and li, and ako between
nieaaurements would be iieeetaary
of AB> Of BF'
i Mork lUinCx 2 p 310
6 In CJ.S ihi un i nnt '■hinm'
, AC = 240 yd., FC = 4S yd.,
often more convenient than that
adjoining dia-
of the top tf
1
1
fiom the groii
il h 1
ci cr tai
the mirror l. 1
'0 ft fi im
the tree
how h gh la th
tree'
7 I rcsters
often detei
mine the
ht kH of -v tie
e by ^n n
stiument
ill i raitetmms Hefei
It Meaa-
m IhMi
m,,lean^
hich tins
initiumtil 1
un liiieted
n ohoun
in thi, di Jijr i n
If the i -Ian
e fiom 4 t
) the foot
ofthotieei bOlf h( =
J iiichts
mdLF =4 n
ht■^ hi 1
1 e 1 tight
of the tree
suit
n fron -1
roimF
\ad shadons cannot be used the
heisht of an objett like a
tiet, or stteplc cin often be
dctLrmined bj a method in
dicated m the d r i w i n g
W hat di'itanee mutt be meas
lied and whv to determine
thi, hught of the tret*
6 'ihow how to find the
h „ht of I tret by placing
I 1 irior m a honzontal posi-
tion on the ground and stand-
ing fco as to see the reflection
if the oijserver's eye is 5i ft.
1 t
, the
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522
r.EOMKTIiV
ACB. On paper make adiaiving of llio triangle ACB to a convenient
scale. On lliis drawing mraanrp l.iic Wm. wliifh represents AB and
hence determine the Icnj^lli of AH.
By use of this method ivc arr> riiivrd Ihn labor of marking out and
measuring the lines CD, CF, and DF.
Apply this rtiethod to the raeasurcraenfc of two objects in your noi^i;h-
borhood which are aepai'ated by an impassable barrier.
In like manner show how the distance from a given point to another
inaecessible place may be dptermined.
9. Also the distance between two place." both of which are inacces-
10. What is the meanuring inst.rument called the diagonal Boale?
How is the principle of similar trianglca tised in it? Show how tile
diagonal seale aiils in the accurate moasurement of lines and hence in
the accurate determination of a distance such as AB in Es. 7, p. 511.
n 11, Id the triangle OAB,
A is a right angle, and 0.1
is 1. By a method which
is Ijeyond the scope of this
book, the length of AB is
computed and found to be
.839 +. Using this tact find
the second triangle,
of the accompanying table, find RQ if angle P is 10°. 20",
Tables ^ving the other sides of all possible right triangles when one
side is unity have been computed and when used as in Exe. 11 and 12,
form the basis of the subjwt of Trigo-
nometry, ity use of this science, after
measuring the length of a single lino a
few miles long on the earth's surface, we
an d t m*n th i' ( n es and laf
po9 1 ons f th pi i-p^ th usands f
mis ana w ihout a. g an
t en ng 1 n By u f tl e« esuU
as a bas th d tan e f th moon
dete m aed aa ipp o\in t t 40 000
miles of the ) SOOOOO n le-
and of th n -e t fa d s( ''0 000
000 Omi 01)0 I
Anglo
AB
08
10
,176
I.OIS
.364
1.064
.Till
1.155
40
.839
1.305
)0
1,192
1,556
()0
1.732
2.000
70
2.747
2.924
5.671
5.759
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PR\rTIC\L APPLICATIONS
A knowledge of these diotances ha^ led to importinl impioi.enii
in method'5 of iixMg,ition and havi, thus fanihtilt^ travel ^nd
merce and increased their benefi''! foe (
usaU
13 On thpdiagramgi\enAJSif,i)Onoii
and the angles as indicated, compute the
length of CD, by use of the table given in
Ex. 13,
14 To the diagram in Ex 13 annex
another tnangle CFD giving it angles %vhi h
are multiples of 10° and compute ll
length of CF
16 A yanloqraph is an in trument fui
drawing 1 phne figun iinuhr to i rimii
plane fagurt It is uaetl for enlai^mg oi -^ uuwu. i>
reducing maps and dra^vmgs It consist-, of four bars, parallel in
pairs and jomted iX c b C and E as shown in the diagram, <i)EC
IS a parallelogram The rods may be joined so that any required
CB'
fixed pivot and pencils are
at b ind F.
Show that A, b, and B are
equals the given ratio
A turns upon a
carried
[Sua, Dra^v a line from .4. to b, and a line from A to B. Prove the
triangle Acb and ACB similar {Art lt^3); henw show that Ab and AB
coincide, etc.]
16 Using the fact that a triangle whose sides are 3 4, and ^ units of
length is a right triangle show how by Btretchmg a 100-ft tape, U>
ponRtruet i right angle is accuiately is posi^ibte (\mong the ancient
Egyptians a class of workmen existed called rope stretchers, whose
busine** was to construct right angleb in this genera! w i\ )
17 The sttongest beam which can be cut from j
gnen round log is found as follows Take AB s
diameter of ihi log and trisect it at C and D Diaw
C E and Dt ± \b n\ meeting the iucumf<rimt \\
h a.tidFrp«puli\(K Draw AF FB BL and If
Pro\e iFBF i rcdinnU also FB U - 1 v^
or iprroxinitih i > 7
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(324 GEOMETRY
|SuQ. _£B is a mean proponional beUveen DB and AB. Therefore
FB' - i AB" (Art^43).] _
In like mnniiM F-V - I AB', ote,]
18. A sphere S weighing JOO ib. rests on ihe inclined plane AB. AC
contains 8 units uf length, BC 6 units. A
force which prevents S from rolling down
the plane would be equivalent to a lifting
force of how many pounds exerted on S and
pai'ailei with l/i''
[Suo. Ile=!ohe the n eight of S (represented
by SP) into two lorces, one perpendicular to
AB and the othei parallel to .-IB. Prove the
triangles ACB and "il'B similar, and obtain
AB: BC = SP: SR, or 10: 6 = 100 lb SR]
The principle involved in this example is of great practical impor-
tance. Thus in many maoliines useful results are often obtained by
representing a force by the diagonal of a rectangle (or parallelogram)
and separating this force into two component forct* represented by
the sides of the rectangle (or parallelogram), onlj one of these com-
ponents being effective. This principle makes poa>ib!e (he action of the
propeller of an aeroplane or steamboat, of the best wat«r wheels and
wind mills, and indeed of all turbine wheels. It also determines the
lifting power of the planes of an aeroplane.
18. A wagon weighing 1800 lb. stands on the side of a hiU which has
a rise of IS ft. for every 100 ft. taken horizontally. Hence what force
must a horse exert to keep such a wagon from running down hill, fric-
tion being neglected?
20. Make up and work a similar example for yourself.
21. Show how the diameter of the earth may be determined by the
following method: Drive three stakes in a level piece of gi'ound (or in a
shallow piece of water) in line, each two successive stakes being a mile
apart, and lot each stake project the same distance above the ground
(or water). By use of a leveling instrument, determine the amount
by which the middle stake projects above a horizontal line connecting
the tops of the end stakes. This distance will be found to be 8 in.
[Sue. Use Art. 343. .\n arc a mile long on the earth's surface may
be taken as equal lo its chord. Then from the diagram of Art. 343 we
obtain the following proportion,
the diameter nf the earth; 1 mi. = 1 mi.; 8 in.]
22. AI?o show that the liistance that the middle stake projects above
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PRACTICAL APPLICATIONS
525
the horizontal liiip connecting the top of the tno end stakes varies as
the squai'e of the dista,nce between the end stakes. Thus if the two end
stakes were placed three times as tar apart !\s in Ex. 21 [that is, 6 miles
apart instead of 2 miles) the bulge of the eai'th between them would
be 3' or 9 times what it was originallj-.
Hence determine the projection of the middle stake (or bulge of the
earth) when the end .sljikos are 4 mi. apart. Also 8 mi. 16 mi, 33 mi.
23, At the seashore an observer whose eye was 10 ft above sea level
observed a distant steamboat whose hull was hidden for a height of
12 ft. above water level by the bulge of the earth. About, how far off
Wiis the steamboat?
24. A seaman in a lookout 42 feet above water level with a glass
could barely see the topsail of a, distant ship, and estimated this top-
sail to be 45 ft. a,bovs sea level. Estimate the distance of the observed
vessel from the seaman.
2E. Work again Exs. T-S, p. 310, Group 56.
36. The moon's distance from the earth's center approximately
equals 60 times the earth's radius. A body failing at the earth's sur-
face goes 193 in. in 1 sec. Hence, if the law of
1, the distance fhe moon falls
gravitation is
;. toward the earth will be
193 ir
Inth.
diagram, let be the earth's center, ABE the
moon's orbit, and CB or AD the distance thi-
moon falls toward the earth iu 1 sec. Taking
the month as 27 da, 7 hr. 43 min. 11 sec,
show that .-ID
193 i n
60=
approximately. This is the ealculation used
by Sir Isaac Newton in testing the truth of the law of gravitation,
27, Any rectangular object, as a book, door, or photograph, is con-
sidered to be of the most artistic shape when its length and breadth
have the same ratio as the segments of a line divided in mean and
extreme ratio (Art. 370). In accordance with this rule, if a window is
6 feet high, how wide should it be?
The division of a line in extreroe and mea
Golden Section. Byraany theGotdenSectio:
tal principle of esthetics, having applicationi
tions of the ideal humas form, in esplanatio
n ratio has been termed the
n is regarded as a f undamen-
! in determining the propor-
n of musical har
. Make up and work nn example similar to Ek, 27.
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•a!
GEOMETRY
EXERCISES. GROUP 91
(Book iV)
The supporting power of a wooden beam (of rectangular cross
n and of given length) varies as the area of the
section multiplied by the heigiit of the beam.
I If the cross section of a given beam is 4° X 8°, com-
e the supporting power of the beiim when it rests
on the narrow edge (4") with il« supporting power
when it regis on its wide edge (8').
2. Two beiims of the same length and material have cross sections
uhich are 2" X 4' and 3° X 8' respectively. Find the ratio of the
supporting power of the two beams,
3. In Ex. 17, p. 533, compare the supporting power of a beam cut
from a log in the manner indicated, with that of a square beam cut
from the same log.
4. Also with that of a beam whose width equals i of the diameter.
5. When an irregular area Uke ABCD i
calculated by means of equidistant offsets, the
following rule is used; To the halt sum of the __
initial and final offsets add the sum of all the S
intermediate offsets, and muUiply tiie sum by the common distance
between the offsets. Prove tiiis rule.
6. Show how the rule of the preceding prob-
, lem could be used to calculate an area whose
re boundary is an irregular curved line.
. Surveyors often determine y
the area of a piece of land, as of ABCD, by taking i,' -
an auxiliary line as NS, measuring the perpendicular
disUnces from A, B, C, D, E to NS, and the inter-
cepts on NS between these perpendiculars, and ^"
combining the areas of the various trapezoids (or j
triangles) formed. Supply probable numbers for the
lengths of lines on the diagram, and compute the ""\ E
area of ABODE. ^
, a d (, (t.. a T ^' '^f^Quently (as when the center of
'~^~j~~~J: i ~ '^e curve cannot be seen from the curve)'
' K'~'--i,^^ a railroad curve is laid out by construe-
^^'^ ting a aeries of equidistant offsets per-
peadicular to the tangent of the curve.
mmm
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'■---.1 ^
"ir
~--4^
^-~-^fi!
■/ "
■"--^ i
PRACTECAL APPLICATIONS 527
Thus, if ABT is a straight track and BC is a Rurve to be laid out tan-
gent tfl AB at B, marli off 6ai, aioi, ajoj, all = d, and construct the _L
offsets a b\ a«ft" Ojfej by Uiing the forniiiU a bu ^ r — Vi^ — n'd'
n here!" IS the radiua of the curve Prove thit this formula is correct.
9 The diagram represent'" tno atiaight parallel railroad tracks
connected bj a compound curve (in this ^ ^
caKe called a crosa-over tta<k) comiostd of p -r*\
t«o tracks with common tangents at D and / 1
b with centers and and having equal -i / ;
ridii Taking the magnitudes as indicated ) /' ]F
oil the figure show that A/ODisareif
angle (use 4rts 122 IfiOl
10 Find u formula tor i m teims of (
d and w (TJae the right tnjnt,U PO.
ind 4rt 401)) \lso tsr ( in tetnib of J^
11. If i« = 4' Hi", d - 10 ft., )■ = 1,^0 ft., i/
find I. ''
12. IE a steamboat is traveling at the rate of 12 miles an hour, and a
boy walks across her deck at right angles to her line of motion at the
rate of 3 miles an hour, draw a diagram to show the direction ot his
resultant motion. From this determine the spi;ed at. which he is going.
13. Make and work a similar example for yourself concerning a mail
bag thrown from a train.
14. Also concerning a breeze blowing into a window of a moving
trolley eai'.
16. Two forces, one of 300 !b., the other of 400 lb., ai-( at right angles
on the same body. Find their resultant.
16. Make up and work a similar example for yourself.
17. A river is flomng at a rat« of 4.25 mi. iier hour, and a man is
rowing at right angles with the current at a rate of 3.75 mi. per hour.
WTiat is the result-ant velocity of the man?
18. If a star has a velocity of 15 mi. a second toward the earth and a
velocity of 20 mi. a second at right angles with a line drawn from the
star to the earth, find the velocity of the star in its own path.
19. A body is moving in the straight line. ID past the point (p. 528).
The distance passed over in a unit of time = AB = BC = CD. When
the body reaches B it is acted on by a force which impels it towaj:d
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GEOMETRY
a diBtancc BF in a unit of 1 ii
!■ bv ;i line draw
by til
Prijvi! lluit 'iio aren swopl over in t!-.
m the body tu will be iinchangc ;
of thf now force. (Thai is, on ilu'
.liagram prove that AOCB =0 AOPB.)
This prinoipie iiccoimts for the fact that the eaiii!
(or another planet or a ooiiiet) moves fuKter in i;,
orbit the neoi-ei' il is l-o the Kim.
20. By traeing tho oiiliine of a map on squanvL
paper, show hoiv to find the aj'pa of an irregular
figure as of some oountry or pari of a country. By
iiBU of this method find the area of aomc part of
the state in which you live.
exercises. group 92
(Book Y)
1. k cooper in fitting a head fo a barrel takes a pair of compasses
and then adjusts them till, when applied six times in succession in the
chine, they will exactly complete fhe eireumferonco. Hr- Ihen takes
the distance between the points of the compasses as Ihe radius of the
head of the barrel. Why is this?
2. Draw a square and convert it into a regidar octagon by cutting
off the corners.
ISuG. Draw the diagonals of the square, bisect their angles of inter-
section, etc.]
3. In heating a house by a hot-air furnace the area of the cross
section of the cold-air box should equal the sum o£ the areas of the pipes
conducting hot air from the furnace. If a given furnace has three hot-
air pipes, each 6' in diameter, and one pipe S" in diameter, and the
width of the coid-air box is 1.5°, how deep should the box be?
i What is the most conMinif nt Tva\ of determining the diameter of
a hot air pipe if you haic no alhpers and the ends of the pipe are not
accessible'
6 ^ half mde running tiack is to hait equal semicit cuUr ends and
parallel straight sides The extreme length of the rectangle together
with the semicircular end*! is to be 1000 fl Find the width of the
rectangle
[Sue Denote the length of the i \d\u> ot ihe emiciriukr ends bj
I and that of one of the parallel idi. straight traj.k=i b-v y and oLtnn
a pair of ■■laiultaDeoua iquafioni |
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PRACTICAL APrLlCATlONfi 529
6, A belt runs over two wheels one ot which has a diameter of 3 ft.
and the other of 6 in. If the first wheel is making 120 revolutions per
minute, how many is the second wheel making. How many revolu-
tions must the first wheel make in order that the second may make
300 revolutions per mtrmte?
7. If in laying a track a rail 10 ft. long is bent through an angle ot
5° 10', what is the radius of the curve?
Assuming {what is not strictly true, owing to friction against the
sides of the pipe, etc.) that the rate of flow through it cylindrical pipe is
proportional to its area of cross section:
8, Work again Ex. 12, p. 277,
9. If a Ij-ii- i'ipG is replaced by a 1-in. pipe, how much i,-- the flow
of water increased?
10. A 3-in. pipe is to be replaced by one which will deliver twice as
mui;h water per minute. Find, to the nearest quarter of an inch, the
diameter of the new pipe.
11. A 2-in. steam pipe conveying steam from the boiler to the radia-
tors in a school building is found to supply only two thirds the needed
amount of steam. What is the smallest even size of pipe that will
convey the needed amount?
12. A city of 40,000 people is barely supplied with water by 12-in.
mains from Ike reservoirs. If these mains are lorn out, and IS-in.
mains substituted, what future population of the city is allowed for?
13. A steel bar I in, m diameter will hold up 50,000 lb. What load
would a bar I in. in diameter carry? What would be the diameter of a
round (cylindrical) bar to carry 150,000 lb.?
14. It Iho center of symmetry of a flat, homogeneous object is the
center of mass, find (he center of mass of a square; of a rectangle;
regular hexagon; circle.
15. It is evident that if the medium of a triangle (BM) be placed on
a knife edge the triangle will balance (for if P!"
be II AC, the pull on P is balanced by the pull on
P'). Hence find the Renter of mass for any triangle.
Tor a regular pentagon.
It is tiscfui to be able Id determine i.he center of
mas-s of an object by neomi^try or by any other
meaus, sincii a knowledge of the center of mas.s ot a Ixidy often
pi\!il)lis us to treat the body in a simple way, tor example, as il llie
l.o(l> iviTc concentrated at a single point.
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5^0 nEOMKTHY
16. if a box has a triaaguiar end, subject to thi
points, at. u'liat single poict oa the end must a
be applied?
17. The cross section of a cylinder is a circle. The woiglit-supporting
strength of a horizontaJ cylindrical beam of given material and length
varies as the area of the cross section times its radius.
Compare the weight-supporting power of two solid horizontal iron
cylindrical beams of the same length and quality of iron, the radii
being 3 in. and 6 in. respectively. (PoinI out and use the short way
of getting the desired result.)
IS. The cross section of a hollow cylinder (i.e. of a tube) is a circular
ring. Denote the outside radius of the tube by R and the inside radius
by r. Then it may be shown that the weight-supporting power of a
hollow cylindrical tube of given length and material varies as the area
. , . iP + r=
of the cross section (ve. of thermg) times — -- — .
It R = i in, and r = 3 in., compare the weighl-wiipporiinir powf^r
of the tube, with that of a solid cylindrical beam of the same length
and same cross sectional area.
19. Make up and work a similar example.
In general a cylindrical tube is stronger than a solid cylindrical beam
of the same length and containing the same amount of material.
Hence in a framework, as in that of an jurship where the maximum
strength must be obttuned from a given amount of material, the metallic
rods and posts iire all tubular. In like manner, bamboo rods, since
theyai'e hollow, are used in an aeroplane instead of solid wooden rods
wherever possible. For the same reasoB, the bones of flying birds, and
many bones in men and animals are hollow and not solid
n 20 To coni^trucf a square which '.hail be ap-
proMm'itely equivalent to a given circle divide
the radiun OA into four equj,! parts and produce
each end of two perpendicular diameters a ell's
; equal to one fouilh of the ra liiis and
rjnnect the extremities of the lines thus formed
slio« that taking the squaie thus foimed aa
equn ilent t the circle i= th( sime as taking
" = 3J, Also find th( per <u!t of erroi in tikmg thi quare aa
equivalent to the circle
21. A short wiy to construct a regular inscribed pentjg n ^.n I aho
il five-pointed stii l w | Liitagrarr ) i-. aa followa Dii \ a ii le and
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PUACTICAL APPLICATIONS
531
two diamelors /IB atiti CD at right angles. Bisect the radius OB at F,
and with f as a center and FC as a radius describe an arc cutting AO
at H. Then CIl is the length of a side of
the regular inscribed pentagon, by joining
whose alternate vertices the pentagram
niay be formed.
As a proof of this, by use ot right tri-
angles, prove
pVio- 2V-5
CH - fl—
C^ee E\ 1" p 301 )
22 f irp iitpra some
lies use the follow mg ^
thod of approximately deteimining the length
t a circumference whose ladiu'i is Inown: Let
O be the giien circle Dran Oi ind OB, radii
t right angles Draw AB j,n1 the radius OELAB
,t D Measure DE Then take cncuraference
^ =b iO + DE
Find the per cent of erroi in this method, taking t = 3 14159-.
23. The following designs are important in architecture or in orna-
mental work (thus the first is a detail in a stained glass window in an
early French cathedral; the second is the plan of the base of a coS-
umn in an English cathedral). Discover a method of constructing each
of tiic following designs, and reproduce each of them in a drawing:
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532 CEOMETRY
21. Construct a square. Take each vertex of the square as a ccater
and one half a side of the square as a radius and describe arcs which
meet. Erase the squai'e and you have a quatrefoi!. By drawing
other circles ajid arcs of circles, elaborate the quatrefoil into an orna-
mental design (see design 11, p. 520).
26. In like manner consli'uct a cinquetoil by use of a regular penta-
gon and develop it into an ornamental design.
26 Troj,t I icgular JicMgon m the same way.
27 4 pa^tmeiif or raouiL may be formed out of regular polygons in
thf, folloivmg '^^^b Shou how to m:i.kc each diagram in the simplest
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APPLICATIONS OF SOLID GEOMETRY TO MECHANICS
AND ENGINEERING
EXERCISES. GROUP 93
(Books \'I and VII)
a the fl:it
1 A carpenti
edge to the surface in various diriK;t;
the flitne^s of a will surface? Wha
these mechanics'
2 EJipHin nhj in object with I
of a surface by applying a ; (might
w does a plasterer test.
ic principle is used by
I'eii legs, as a stool or tripod,
object with tour legs, iis a table,
ever use foiar-leggcd pieces .of
furniture?
3. How can a earpt;nter
get a comer post of a house
in a vertical position by use of
a carpenter's Btiuaro? What
geometrical prineiple does he
4. The diagram is the plan
ot a hip roof. The slope of
each face of the roof is 30°.
Find ihe ienglh of a hip rafter
sAB.
1 (he pli
trianRle DBC repre,wnli
Hence it may l>e fih
El section ot the roof at
20
I ISC = - v;!. In
like manner by taking a section through
BF, it is found that AC = 10. Hence in
%e triangle ABC, AB may be found.]
6. Find the area of the entire roof
represented in Ex, 4,
6. Make drawing showing at- wiiat
fttigle Ihe two ends of a rafter like liC in Ex. 4
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5^4 GEOMETRY
7. Make drawings ?liowinR at what angle a jack raftpr like 12 in
Ex. 4 rauBt be cut.
ISuiJ. To determine how the end 1 of the jaek rafter must be eut
use the principle that two intersecting straight fines determine a plane
(Art. 501). The cutting plane at 1 must make an angle ;it ihe side
of the jack rafter equal to angle CBH, and on the top of the jiick rafter
equal to angle ABC]
8. What is a gambrtjl root? Make up a set of examples coticorning
a gambrel roof similar to Exs. 4-7.
9. By use of Art. 615, show that a page of this book held at twice
the distance of another page from the same lamp rei'ei\'es one fourth
the light the first page receives.
10. The supporting power of a wooden beam varies directly as the
area of the erosa Heotion times the height of the beam and inversely as
the length of the beam. Compare the supporting power of a beam
X2 ft. long, 3 in. wide, and 6 in. high with that of a. beam 18 ft. long, 4 in,
wide, and 10 in. high. -Uso compare the volumes of the two beams.
EXERCISES. GROUP 94
(Books VIII and IX)
1. A hollow cylinder whose inside diameter is 6 in, ia partly filled
with water. An irregularly shaped piece of ore when placed in the
water causes the top surface of the water to rise 3.4 in, in the cylinder.
Find the volume of the ore.
2. What is a tubular boiler? What is the advantage in using a
tubular boiler as compared with a plain cylindrical boiler? If a tubular
boiler is IS ft. long and contains 32 tubes each 3 in. in diameter, how
much more heating surface has it than a plain cylindrical boiler of the
same length and 36 in. in diameter? (Indicate both the long method and
the short method of making this computation and use the short method.)
3. If a bridge is to have its linear dimensions lOOO times as
great as those of a given model, the bridge will be how many times
as heavy as the model?
Why, then, may a bridge be planned so that in the model it will sup-
port relatively heavy weights, yet when constructed according to the
model, falls to pieces of its own weight?
Show that this principle applies to other ronstructions, such as
buildings, machines, etc., as well as to bridges.
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rRACTICAL APPLICATIONS 535
i. Work again Eks, 23-25, 27, p. 473.
6. Make and work for yourself an example similar to Ek. 24, p. 473.
6. SouniJ spreads from a center in the form of tlie surface of an
expanding s[)here. At the distance of 10 yd. from the source, how will
the surface of this sphere compare with its surface as it was at 1 yd.?
How, then, does the intensity of sound at 10 yd. from the source com-
pare with its intensity at a distance of 1 yd.'f
Does this law apply to all forces which radiate or act from a center
aa to light, heat, magnetism, and gravitation? Why is it called the
law of inverse squai'es?
7, If a body be placed within a spherical shell, the attractive forces
exerted upon the body by different parts of the shell will balance or
caned each other. Hence a body inside the earth, as at, the foot of a
mine, ia attracted effectively only by the sphere of matter whose radius
is the distance from the center of the earth to the given body. Hence,
prove that the weight of a body below the surface of the earth varies
as the distance of the body from the center of the earth.
[Stjg. If W denote the weight of the body at the surface, and w its
weight when below, R the radius of the earth, and r the distance of the
body from the center when below
the surface, show that
8 The heht ol tl e sun falling
on a smaller sphere, as on the earth oi the moon cau-,es that b^dv to
cast a conical shadow. Denoting the radium of the sun bj R the
radius of the smaller sphere by r, the diatanct between the two spheres
dr
by d, and the length of the shadow by I, show that I = - _ ■.
Find I when d - 92,800,000 mi,, R = 433,000 mi., r = 4000 mi.
9. If in a lunar ecli|}se the moon's center should pass through the
axis of the conical shadow, and the moon is traveling at the rate of
2100 mi. an hour, how long would the total eclipse of the moon last?
How long if the moon's center pas.?ed through the earth's conical
ahadon- at n distance of 1000 mi. from the axia of tlie cone?
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531; c i)':oMr,'ri!v
10, If the moon's diaraotcr is 2160 mi., finil Ihn length o( the moon's
shadow us caused by sunliglit.
11. If the distance of the moon from the earth's eenter VMrics from
221,600 mi. to 232,970, show bow this explains why some ei;lipscs of the
mm are total and others annular.
Why, also, at a given point on the earth's surface is an eclipse of the
sun a so much rarer sight than an eclipse of the moon? Why, also, is its
duration so much briefer?
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FOEMi.n.AM OF" PLAICE OEOMETEY
SYMBOLS
6, <!=siiles of Uianglo J/tr. /,'--^iadiH3 of a tilrcle.
s = i{a + b + c). i).-diameter of acirele.
ftf = aUitude on sklo r. (' = cirpiimferenee of acirele.
5ii( = meiliiin on side ('. f = i'iitlius of an inso'lbed
*c=biaettorotanglB.>i.!"Hile pirole.
SLiii.' .-. T = -r-apprax. (or :i. 141(3— ).
id)n=liiiB segnjouLs. K=area.
/■= [loi'iuietcr. /i = liase of a triangle.
6\i = slde of a regular polyi^oii h =altitude of a triangle,
ol » sides. ''[ anil ?i- = bases of a trapezoid.
LENGTHS OF LINES
1. In a right triangle, C being the tight angle,
f' = n' + &'. Art. 346.
'2. In a riglit, triangle, ( and m being Ihe projections of a and fe
f, and li, tho altitude on c,
3. In an oblique triangle, m being tho projectioo of b on e,
if aisopposite an ai^^ito Z, a' = b= + c= — 2 cX m. Art. 319.
if a is opposite an obtuse Z, a' = 6' + e' + 2 cX*"- Art. 350.
».-iJ/<«-<.)(--»)l— «J
Art. 393.
me=iT/at«^ + 6"j-c^
Art. 353.
If I and m are the segments of
opposite, a:J' = l :
Art. 3li3„
c made by the biaeotor o( the
m. Arts, 333, 336.
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ISDEX OF DEFiXlTIOXS ASD FOEJIULAS
^bbrp It on
ilgebr ic analy 3
Algebra e u eti o
■ilternat on
Altitude of eone
of frustum f
of frustum of p
of parallelogr n
otpr.m
of pyrom d
of spher c I
of trapt
of tr ngle
of zone
Anil J olu o
An{,!c
at CD
to
diled
formed hv a
etra ght I n
formed bv t
inscr bed n
I
Angle of lull
poljhedrnl
right
salient ,
sides of .
splierieal
straight
tetraliedial
trihedinl
vertex ot
Angles, adjacL'i
alteniato-iiili
llOH)ologOll;^ .
interior , .
of polygon
of quadriliiti
of spherical
Appolom
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IND]:X OF IH'.FINITIOXS AXD FOTiMlXAS
major , .
Archimedes
Arcs, eonjligiite
Area ol surface
Aryabhatta
August, E. F.
Auxiliary linea
Axioms, generiil ,
geometric
Axis of circle of splie
of circular cone
of regular pyiaoiii;
of symmetry
Babylonians .
Base of cone .
of isosceles triai
of pyramid
of spherical pyr;
of spherical sect
of triangle ,
! of <
lulcr
of fnislum of or
of frustum of ]ii
of parallelogram
of prism
of spherical segment .
of trapezoid
of zone . .
Bisector of triangle .
Bodies, the Three Round
Center of circle .
of regular polygon
of sphere ....
-'■'
Chord
Circie 10
are of
center of ... .
circumference ul". tormiil:
for
circumscribed . . .
formula for area of
103
,406
103
103
103
274
103
103
inscribed ....
radius of ... ,
sector of ... ,
small
Circles, conecnliic . .
escribed ....
105
10.3
104
427
ID.i
rireum-centcr of Irmn^'le
Civcumtercncp . . . IJ
Classification of ]>islvln-dion
of triangUM . . .
Complement ....
Composition . , , .
81
.103
3G0
2,33
125
16
181
Conclusion ....
Coninirrent lines .
24
80
Cone
altitude of ...
^^xis of
base Of
circular
circular, fotiiiuhi for vo
uine of ... .
circular, formulas for la
eral and total area .
412
412
413
413
418
417
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INDEX OF DEFINITIONS AXD FORJIULAS
Cone, elements of
lateral surface of .
oblique circular
of revolution .
right circular .
vertex of
Cones, similar
Congruent figures
Conical surface .
directrix of
element of .
generatrix of .
vertex of . .
Conjugate area
Conaequenta
Constant .
geometrical .
Constants .
Continued proportioi
Continuity, principle of .
Converse of a th<
Corollary .
Cube . . .
Curved spaces
Curved surface
Cylinder ,
altitude of .
bases of
circular .
circular propei
elements of .
lateral surface of .
oblique
of revolution
of revolution, formulas fc
lateral and total arc;
of . . .
right . . .
right ciroula.
Cylinder, right
Cylinders of re'
Cylindrical sut
elor
t of .
D t mn d
Pl
D g n 1
pol
K
f p lyl
J
f q d
1 t
D m t
1
t ph
Dh d 1
1
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I^■DEX OF DEFIXmOXS -VXD FOIi:\IUIA
Duality, Fi^
■ipk of
Edge o£ dihedral ni,
Edges of polyhedral
of polyhedron .
Egj-ptiiins .
, . 400, 401, 432,
Elements of conical f.
of cylindrical surface .
Enunciation, general
pftiticular
Equal figures .
Equivalent figures .
solids ....
Euclid
Eudosua . . .493
Euler
European ....
Extreme and mean ral
Extremes ....
Faces of dihedral angle ,
of poiyhedTiil angle
of polylirdmn .
Figure, curviliiiuar
geometrical .
mixtilinear .
plane_ . . .
Figures, roctilinoar
Formulas of Piano Geo net
for lengths of 1 ne j
of Plane Oeoniet fo
areas of plane fi re^
of Solid Geometry fc
of Solid Geometry ft
volumes ,
Foot o£ perpendicular
Fourth propoilionul
PruBtuni of toiiu.
altitude of . ,
bases of . . .
formula for l.ileral
of ....
formula for volume i
lateral surfacp of ,
slant height of
Frustum of pvraniid
altitude of . . .
base
of
slant height of
;al !
of c.rlintirical sui
Geometric solid ,
Geometry .
epochs in development of 4
history of .
modern .
Non -Euclidean ,
Jrigu
of
e ka
400
102, 403, 494
Ha n on i
. , . 108
Hept oon
.... 74
He o
. . .496
He 1 on
... 74
He h i on
. . .360
H n loos
400, 432, 497
Hpp U3
, . .408
H ppoc atus
491, 492, 406
Hstor ofg
t
490-498
Honol go 3
a =1
.... 34
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IXDEX OF DEFINITIONS AXD FOKJIULAS
Homology, Principle ot
487
I ne d ided tein^llv
Hyperapafie ....
48S 1
ot centres
Hypoteniiae ....
33
I rallel to pi
Hypothesis ....
. 24
I rj end eul r t j 1
Icosahedron ....
, 3G0
°
In-center of triiinplE!
. s;
L a au 1 irv
Inclination of line to pia
e 347
concurre t
. 1-25
Lobatche 1
ratio
. 1:25
Loci^
Inference, immediate .
. 23
Log eil U t)l
Inversion
. 180
Lune
Isoperi metric figures
. 280
inn-lp ot
for i f
Jews
, 43-
apler al degr es
torn nU for or a
Lateral nrea of frustum
ol
square un ts ot a
pyramid, formula for
. 379
ot pyramid ....
. 37
Alagn tude n
Lateral edges of prism
3G1
mcommen 1 1
of pyramid
377
Alaji nun
Lateral faees of prism
ill
"Mean p opo t a 1
of pjramid
3"
Mea
Lateral surfaoe uf cnne
41'
Me h n cal n et} 1
o£ ej Under
401
M d an of trapezo 1
of frustum ot eon-^
414
ot tranje
Legs of iso.^d<^luaa>
3
"\Ienela s
of riplit trnn„-lL
31
Method ilgebi
of trapwoid
Iinut
(1
\1
of 1 ts
Method loj, c 1
Limits, metiiol of
\1
meeh i -x]
Line
1
ot n I
broken
11
tion«
curbed
n
hcto e 1
divided exttnmllj
1')^
Metr e sjste
dnided hatmonii ,\h
I'lS
^\ nor n <■ ot <■ 1
di\ided in p\titiiic i
ii
■\I te
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INDEX OF DKriMTiONS AN)") FOILMULAS
Nappes of cono .
Negative quaniili™
Non-Eup]idean gcoi
Numerical t
Octagon
Octahedron
Opposite of a thci
Origin of geoiiK'tr
Ortlio-eenter
Parallei lines
Parallelogram
altitude of .
bases of
Parallelopiped
right . .
rectangular
Parallel plants
Pentagon .
Pentedecagon .
Perimeter of p'll
of triangle .
Perpendicular liii
planes
Pi {«) . .
Plane . . .
determination
Pianes, parallel
Plato . . ,
Point . . .
PointB, coney lie
Polar distance of
triangle , ,
Poljhediai angles, equal
synimetTical
Polyhedron
diagonal of
edges of ,
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IKDEX OF DEFINITIONS AND FOT!ML"LAS
Polyhedron 1 1
Polyhedrons
classificiti n
Postulnfi.
of solid gpon
Postulites
logical
Prism
altituclc if
base
of
oirpumsLribed aboi
inscribed in oil!
lateral area o!
lateral edges of
lateral fates of
oblique
quailrdnjrular
right
Tight sectirn i
trianguhr
tmncuteJ
Pri'imaioid
bases of .
formula for voli
Prismoid .
Problem
Pi ejection of lint
of line on plan
of point on line
of point on plar
Proof
by auperpciiiio i
forms of
method? 1)1
proportion
eontinu.d , .
torins of . .
third
axi ,f
base of
circumscriljei about
fiustum of
jnbcniel in on
latcnl ed„ ol
lalor
fie
qmdr nfnil r
regular
regular slint 1 tif,!
spherical
trnngul r
truncated
verte\ of
PTtha'
491, 492, 4!)S, 4!i7
Q d t f
Qu d 1 t 1
angl f
n ! tud
II3 p i>ortionaI
1 p neipic of
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IXDEX OF DEFINIT10^:S AND FORMLXAS
Rhombus
(it)
f.o!id^ phTBin!
Right section of <L hnloT
4f)5
Spaei-- cuned
of pusm
3bl
Sphere
romaiis 4
4 4')0
ftilS of
Round liiijic Tl t n rep
a-^s
center of
Eeholium
23
Lircuniscrihtd \bna
Vp 1 t line
104
hedron
Bfond
17
diameter of
Sect
14
formula fot arpx
Election o! I 1 ilJ un
J1)0
face of
Sector of cink
104
formula for ^ohlnl
of cirplo fori lU fjr
great circle of
of spilt re
4bl
inscribed in polili
Sectors amiJ.r
275
poles of
Segrapnt of i tlo
114
radma of
of line
14
small circle of
of sphere
4b2
Spherical angle
Segments un ilar
2T3
degrees
Semicircle
101
Seniicireumfeceii"e
103
SpheriPil poh„ua
''ides of angle
14
angles of
of poljgon
73
sides of
of quadiilateril
u6
vertices of
of triangk
32
Sphericil piramid
Similar cones of r oluti
on 113
base of
eUmdeis oi i ol li ii
404
vertex nf
polvgina
(i3
130
Spherical sector
base of
pohhedrons
J03
formula for lolum
Btttors of circlps
Spherical segment
segments of civ(1p9
2-j
altitude of
Slant height of coik
413
of frustum of cone
414
formulis tor loluu
of frustum of pyramid
of one bnie
of regular pyr-iinid
^77
Spherical triangle
Solid
bi- rectangular
geometry ....
18,319
formula for area
Solids, equivalent . .
. 303
sqnnre unifs of
geometric ....
. 11
tri-rpcfangular
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INDEX OP DEFINITIONS AND FOEilULA;
Spheiicil trim,
mtntal
Bymmetrici 1
Spheneal wedgt.
Spherid
Stiaight line
Superposition
Supplement
area of
conical
Tangent circles
langent conimon evternil
Ime to tpleie
plane to cjlinJer
plane to &phei
Terms of a proportioi
Tetrahedral angle
Tetrahedron .
Thales . ,
Theorem
Theory of liniil
Third proport:
Transversal
Trapezium
Trapezoid .
altitude of .
Iriangl
d' 7-1
acute
33
altituilts jt
33
base of
33
bisectors of
34
bisectois of forniU
for 213
clagaifications ut
3', 33
equiangular
33
equilateral
S2
isosceles
32
medians of
34
oltuse
33
pohr
441
ii^ht
33
icilcne
32
spheueal
438
\crfc\ angle of
33
\eitex of
33
riihedral angle
349
bi rectangulir
330
isosceles
350
lectanguUr
350
tiirtotangular
350
Ungula ....
. dlil
Unit of measure . -
of surface . . .
. 231
of volume . . .
, . 3S3
Units of angle . .
, 17
of spherical surface
. 452
Variable ....
. . 120
Vertex ani^le .
. . 33
Vertex of angle . .
. . 14
of polyhedral angle
. , 349
of pyramid . ,
. . 377
of spherical pyramid
, 461
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IXDliX OF nEFIXITlOXS AND FORMULAS
Vut'teic of triangle .
Vertices of polygon .
of polyhedron .
of quadrilateral
of spherical polygo
of triangle .
Volume of solid .
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